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Bonk > WS S 

Copyright N°. /?a a 


COPYRIGHT DEPOSE 
















A LABORATORY AND CLASS-ROOM 
GDIDE TO 

QUALITATIVE 
CHEMICAL ANALYSIS 


BY 

GEORGE F. WHITE, Ph.D. 

Associate Professor of Chemistry in Clark College 
Instructor, Clark University 


SECOND EDITION 
REVISED AND ENLARGED 



NEW YORK 

D. VAN NOSTRAND COMPANY 

Eight Warren Street 

1920 

J /y o 





Copyright, 1916, by 
D. VAN NOSTRAND COMPANY 


Copyright, 1920, by 
D. VAN NOSTRAND COMPANY 


OCT 30 1920 


PRESS OF 

BRAUNWORTH & CO. 
BOOK MANUFACTURERS 
BROOKLYN. N. Y, 

©Cl,4601286 



PREFACE TO SECOND EDITION 


The book has been carefully revised and brought up to 
date. New matter has been added, in particular a discussion 
of spectrum analysis; it is again emphasized that such a brief 
discussion is intended to draw the attention of the student to 
a subject, further details of which must be acquired by con¬ 
sulting larger works. 

The author takes especial pleasure in acknowledging the 
courtesy accorded him by the publishers of this book. 

G. F. W. 


Clark University, 
August i, 1920 





PREFACE TO FIRST EDITION 


In compiling this book, the author has endeavored con¬ 
tinually to bear in mind the fact that the. main objective 
of a course in qualitative analysis is the attainment of a 
knowledge of the fundamental principles underlying the dif¬ 
ferential behavior of elements and compounds. Ability to 
devise and execute schemes of analysis should be developed 
naturally. With this point of view, a study of reactions has 
been combined with methods of analysis, the latter being 
considered as practical applications of previously observed 
phenomena. It is considered that this book satisfies the need 
of a manual for students who have completed the usual 
college course in general chemistry. Such students have 
acquired some knowledge of the physical laws underlying 
chemical reactions and yet, in the author’s experience, too 
often have only very indefinite conceptions of these laws. 
There are books at hand which present advanced theories 
in detail, and there are others which lay especial stress 
on analytical procedures and laboratory practice. Such texts 
are admirable for reference. This book is offered as a work¬ 
ing manual which presents the essentials of both theory and 
practice, but which also suggests the possibilities for more 
extended study and experimentation. 

Emphasis has been laid in the first part of the course 
on the reversibility of reactions, equilibrium phenomena, and 
the laws governing the behavior of electrolytes, while many 
applications of these principles have been left to the student 


VI 


PREFACE 


in his later work. The brief outline of the electrolytic dis¬ 
sociation theory presented, is meant to be supplemented by 
class-room questions .and instruction; this is also true of the 
individual base-forming and acid-forming constituents, analyt¬ 
ical procedures, etc. 

While much may be said for the progressive study of the 
base-forming constituents beginning with the alkalies, the 
silver-group metals have been first considered in order to 
co-ordinate the study of reactions with the scheme of analysis 
offered. The first year’s work in general chemistry should 
have furnished the student sufficient knowledge of the alkali 
group and the general properties of the metals to enable 
him to appreciate the full significance of the reactions as 
carried out. 

The dimethyl-glyoxime test for nickel, and the nitroso- 
beta-naphthol test for cobalt have been found in this labora¬ 
tory to be particularly successful in the hands of the beginner. 
The cobalt test was successful, however, only after being 
modified in such a way that there is no possibility for the 
reagent to be precipitated. 

The author greatly appreciates the suggestions of Dr. 
Frederic Bonnet, Jr., Worcester Polytechnic Institute; Mr. 
John T. Ward, Clark University; and the particularly helpful 
advice and aid of Dr. B. S. Merigold, Clark College. 

George F. White. 

Worcester, Mass., 

July i, 1916 


CONTENTS 


PART I 

STUDY OF REACTIONS AND ANALYTICAL PROCEDURES 

PAGE 

Introduction. i 

Theories of Solutions. 2 

General Instructions. 26 

Reactions of the Base-forming Constituents and Basic Analysis. ... 27 

Reactions of the Acid-forming Constituents. 99 

PART II 

SYSTEMATIC ANALYSIS OF UNKNOWN SUBSTANCES 

Preliminary Tests. 127 

Analysis of a Solution. 131 

Analysis of a Solid Substance for the Base-forming Constituents. .. 133 
Analysis of a Solid Substance for the Acid-forming Constituents.. . 142 

APPENDIX 

Analytical Tables. 147 

Directions for the Preparation of Reagents and Test Solutions.... 155 

Solubilities of Difficultly Soluble Compounds in Water. 160 

Relative Solubilities in Water and Acids. 161 

Percentage Ionization of Acids, Bases, and Salts. 165 

Table of the Elements Arranged According to the Periodic System 166 

Table of Atomic Weights. 167 

vii 



















































































































QUALITATIVE CHEMICAL ANALYSIS 


PART I 

STUDY OF REACTIONS AND ANALYTICAL 
PROCEDURES 


INTRODUCTION 

1 . Qualitative chemical analysis is the science by means 
of which the elements and chemical compounds of which a 
substance may be composed are identified. By this defini¬ 
tion we therefore include the determination of the nature 
of all inorganic and organic .substances—rare metals and 
their compounds, thousands of carbon compounds, foods, 
drugs, oils, fertilizers, and so on. The field must necessarily 
be limited for an introductory course; only the more com¬ 
mon elements and inorganic compounds (with few exceptions) 
are considered in this book. 1 

2 . Theoretically, qualitative analysis is sharply distin¬ 
guished from quantitative analysis, which latter science deals 
with methods for ascertaining the actual amounts of con- 

1 For a systematic procedure for both the common and rare inorganic ele¬ 
ments and compounds, the student should study the articles of A. A. Noyes 
and co-workers. J. Amer. Chem. Soc., 29, 137 (i 9 ° 7 ); 30, 481 (1908); 31, 
611 (1909); 34, 609 (1912). 





2 


QUALITATIVE CHEMICAL ANALYSIS 


stituents present in any given substance. A quantitative 
analysis should consequently follow the identification of these 
constituents. In actual practice, however, a good qualita¬ 
tive analyst is able to estimate, roughly, the relative propor¬ 
tions of constituents. By suitable attention to the quantity 
of unknown substance used, and to the products yielding 
indications of the original constituents, and by care and 
neatness in manipulation, the above object may be attained. 

3 . The methods of qualitative analysis are such that an 
unknown substance is caused to enter into chemical reactions 
by the use of various reagents; the base-forming and acid¬ 
forming constituents (elements, or radicals in a compound) 
are distinguished by the character of the products formed 
in the reactions. These products serve to identify the original 
constituents by certain properties perceptible to the senses, 
such as color, odor, state of aggregation, etc. As most of the 
reactions carried out in qualitative chemical analysis take place 
in aqueous solution, the nature of such solutions will be 
briefly discussed. 

THEORIES OF AQUEOUS SOLUTIONS 1 

4 . In order to understand certain analogies between gas mixtures 
and aqueous solutions, the student should recall the laws of gases, 
which are summarized below. 

Boyle’s Law: pv=constant. The pressure p and the volume v of 
an isolated quantity of gas are inversely proportional to each other 
if the temperature is kept constant. 

Gay-Lussac’s Law: pv=p 0 v 0 (i+ 1/273/). p 0 and v 0 are the pressure 
and volume respectively, of the gas at o° C.; / is the temperature in degrees 

1 It is assumed that in the first-year course in chemistry the student has 
become familiar with the general theories of gases and solutions, at least with 
the essentials of the phenomena concerned. Only those theories particularly 
applicable to qualitative analysis will be emphasized here. 


THEORIES OF AQUEOUS SOLUTIONS 


3 


centigrade at which p and v are measured. In words, the pressure- 
volume product of a gas changes by 2+3 of its value at o° for each 
degree difference in temperature. At constant volume, the pressure 
will increase °f its value at o° for each degree rise in temperature; 
or at constant pressure, the volume will increase 3 of its value at 
o° for each degree rise in temperature. If we express the temperature 
in absolute degrees (absolute zero is 273 0 below o° C. and at that point 
the above law would indicate that the volume of the gas would be zero) 
we have: pv=p 0 v 0 T/2y^, where T is the absolute temperature. 

Dalton’s Law. In a mixture of gases exerting neither physical nor 
chemical interaction, each gas exerts the same pressure as if it were 
alone present in the same volume. The total pressure of the mixture 
is equal to the sum of the partial pressures of the components. 

Avogadro’s Hypothesis. Equal volumes of all gases at the same 
temperature and pressure contain the same number of molecules. If 
we consider one gram molecule of a gas (the molecular weight in grams, 
sometimes called a mol), ^0/273 will have the same value for all gases; 
it is a constant which we express by R, and pv=RT. The so-called 
gas constant R is independent of the nature of the gas. 

If the pressure is measured in atmospheres and the volume in liters, 
the volume of one mol of a gas at a pressure of one atmosphere is 
22.4 liters at o° C. (or the pressure of one mol of gas of volume one 
liter, is 22.4 atmospheres at o° C.), and pv—22.^/ 273T=0.082T. R, 
therefore, is equal to 0.082 liter-atmosphere. 

5. Density and Molecular Weight of Gases. By Avoga¬ 
dro’s hypothesis we assume that there are the same number 
of molecules in equal volumes of all gases at the same tem¬ 
perature and pressure. Vapor density determinations, there¬ 
fore, have been used to establish the molecular weights of 
elements t and compounds, since the relative densities of gases 
under the same conditions of pressure and temperature 
would be a measure of the relative weights of the molecules. 

Thus the vapor of chloroform (which is a compound of carbon, 
hydrogen, and chlorine) has been found to be 59.7 times as heavy as 
an equal volume of hydrogen at the same temperature and pressure. 
Since the molecular weight of hydrogen is 2.0, the molecular weight of 
chloroform is 119.4. Hence its molecular formula is CHCI3. 


4 


QUALITATIVE CHEMICAL ANALYSIS 


Soon after the application of the hypothesis apparent excep¬ 
tions were noted, molecular weights being obtained in some 
instances which were too low to correspond with the well- 
known chemical behavior. For instance, the molecular weight 
of gaseous ammonium chloride, as determined by this method, 
is very much less than that calculated from the formula 
NH4CI. But ammonium chloride vapor can be shown to 
consist largely of hydrogen chloride and ammonia—in other 
words the compound is broken up into smaller molecules. 
It is dissociated. This dissociation is greater as the tempera¬ 
ture is raised, and it causes a decrease in density and in the 
calculated molecular weight. The dissociation is reversible , 
since the cooled vapor consists of ammonium chloride, and 
we may write the reaction for the effect of heat on ammo¬ 
nium chloride thus: 


NH4CI *=> NH3+HCI. 

Such dissociation may be visually observed in the case of 
phosphorus pentachloride: 

PC 1 5 ^ PCI3 + CI2, 

where the greenish-yellow color of the chlorine is seen in the 
heated vapor. The color fades out when the system is cooled. 

If the molecular weight of phosphorus pentachloride were determined 
by the vapor density method, 208.3 would not be obtained, as the for¬ 
mula PC 1 5 would require. At a sufficiently high temperature, where the 
dissociation into two molecules—one of phosphorus trichloride and one 
of chlorine—is complete, one-half of the true molecular weight, or 
104.2, would be obtained. 

The decrease in molecular weight of phosphorus pentachloride with 
rising temperature may be observed from the following experimentally 
obtained values: 


THEORIES OF AQUEOUS SOLUTIONS 


5 


Temp, in 
Degrees C. 

182. . . 
200 . . . 
250. . . 
300.. . 


Molecular 

Weight 

. ..147 
. . . 140 
• • .115 
...106 


Another example of gaseous dissociation is that of water vapor> 
which dissociates into hydrogen and oxygen at high temperatures: 

2H2O 2H2 | 02 « 


As the temperature is raised the dissociation is increased. If the 
temperature falls water vapor is reformed. The percentage dissociation 
at different temperatures is given below: 

Temperature. 1500° 2000° 2500° 

Percentage dissociation. 0.02 0.59 3.98 

Thus at 2000°, 2 mols of water vapor or 36.03 grams, will dissociate 
and yield 2 —(2 Xo.oo59) = 1.988 mols or 35.82 grams of water vapor, 
0.012 mol or 0.024 gram of hydrogen, and 0.006 mol or 0.19 gram of 
oxygen. 

6. Osmotic Pressure and the Molecular Weight of Dissolved 
Substances. If a crystal of copper sulphate is dropped 
into a tall beaker of water which is kept free from any dis¬ 
turbance such as jarring, as the crystal dissolves, the color 
will gradually spread from the bottom to the top of the 
liquid in the beaker (several days or weeks will be required). 
The salt has the power of difusion and the process is analo¬ 
gous to the diffusion in gases, where a gas will distribute 
itself uniformly throughout a space fdled with another gas. 

The analogy may not be carried too far, since a gas will diffuse 
into an empty space, whereas the crystal of copper sulphate requires 
a solvent for diffusion. 


If the diffusion of gases is impeded by a semipermeable par¬ 
tition (a partition that allows the passage of only one gas) a 
pressure will be exerted on the partition by the gas incapable 








G 


QUALITATIVE CHEMICAL ANALYSIS 


of permeating it. Thus, if a mixture of nitrogen and hydro¬ 
gen is placed in a heated palladium vessel which is itself 
enclosed in an evacuated space, hydrogen will pass from the 
inner to the outer vessel, since palladium is permeable to it. 
The partial pressure of the hydrogen inside the palladium 
vessel will become equal to the pressure of the hydrogen 
outside, since an equilibrium will be set up from the tend¬ 
ency of the gas to diffuse uniformly through space. There 
will, however, be an excess of pressure in the inner vessel, 
since the nitrogen cannot pass out, and its partial pressure 
will be added to that of the hydrogen. 

A pressure can be demonstrated in aqueous solutions, and 
it can be shown to be analogous to the gas pressure of the 
nitrogen in the above experiment. Thus a properly prepared 
clay vessel in whose pores has been deposited a film of 
copper ferrocyanide will allow the passage of water, but not 
of glucose—it is a semipermeable membrane. If we place 
an aqueous solution of glucose in such a closed vessel con¬ 
nected with a manometer and placed in a larger vessel con¬ 
taining only water, water will pass from the outer vessel 
through the membrane into the inner vessel. The pure 
solvent tends to dilute the solution. A pressure will be 
evidenced over the solution and can be measured in the 
manometer. The phenomenon is connected with the power 
of the dissolved glucose to diffuse and to distribute itself 
until the entire system is of uniform concentration. Since 
the glucose cannot pass through the membrane, there is a 
bombardment of its molecules against this membrane. The 
pressure in a solution causing this bombardment and made 
evident by separating the solution from the pure solvent by 
a semipermeable membrane is called the osmotic pressure of 


THEORIES OF AQUEOUS SOLUTIONS 


7 

the solution. Osmotic pressure, then, may be defined as the 
excess pressure in a solution over that of the pure solvent 
when the two are separated by a semipermeable membrane. 

7. Now a remarkable analogy has been discovered be¬ 
tween the laws of gaseous pressure and of osmotic pressure of 
solutions. Osmotic pressure has been shown to obey the laws 
of Boyle, Gay-Lussac, and Dalton, if for the gaseous pressure 
we substitute osmotic pressure, and for the gaseous volume the 
volume of solution in which the osmotic pressure is exerted. 

Van ’t Hoff drew the above analogies and also extended 
the application of Avogadro’s hypothesis to solutions. He 
stated that the osmotic pressure of a dissolved substance is 
equal to the pressure which that same substance would 
exert in the gaseous form at the same temperature and con¬ 
centration. We may say, then, that equal volumes of solu¬ 
tions of the same osmotic pressure and at the same temperature 
have an identical number of dissolved molecules. Molecular 
weights of dissolved substances may, therefore, be measured 
by comparing the amounts of these substances with an 
amount of substance of known molecular weight dissolved in 
the same volume of solution and producing the same osmotic 
pressure at the same temperature. But again discrepancies 
were noted, as in the application of Avogadro’s hypothesis 
to the molecular weight determinations in gases. While 
the molecular weights of many substances such as glucose, 
C 6 Hi 2 0 g, and ethyl alcohol, C2 HgO, dissolved in water, are 
normal—that is, give values which conform to the known 
chemical behavior and are the same as those determined by 
the vapor density of the gaseous substance (decomposition 
prevents the determination of the vapor density of glucose)— 
there are many exceptions. In fact, the substances most 


8 


QUALITATIVE CHEMICAL ANALYSIS 

frequently met with in inorganic chemistry give low molec¬ 
ular weights as measured by the osmotic pressure method. 
Arrhenius explained this important behavior so simply and 
clearly that his theory, outlined below, has been practically 
universally adopted. 

The molecular weight of alcohol, C 2 H.O, is found to be 46.0, both 
by the vapor density method and by osmotic pressure measurements 
when dissolved in water. For sodium chloride, NaCl, dissolved in water, 
58.5 will not be obtained as the molecular weight by osmotic pressure 
measurements; but, depending on the dilution, a value less than this 
and approaching 29.3 will be found. That is, it will take only 29.3 
grams of sodium chloride to produce the same osmotic pressure as 46.0 
grams of alcohol (whose molecular weight is normal) in the same vol¬ 
ume of solution and at the same temperature. Instead of 58.5 grams 
of NaCl as we would expect from the known molecular formula of the 
substance, only one-half as much is required, indicating the presence 
of twice as many particles or molecules in this solution as in the alcohol 
solution (osmotic pressure, according to the application of Boyle’s law 
to solutions, is proportional to the concentration of the solution). 

8 . Ionization Theory. From the behavior of vaporized 
ammonium chloride we might assume by analogy that those 
substances which suggest low molecular weights in solution 
are likewise dissociated into smaller molecules. Arrhenius’ 
theory involves this assumption, but goes further. He saw 
that those solutions which give normal molecular weights 
for the dissolved substances do not conduct the electric 
current, but solutions which give abnormally low molecular 
weights are conductors of the electric current. Such sub¬ 
stances, which in solution conduct electricity, are called 
electrolytes. There must be charged particles in these solu¬ 
tions, capable of transporting the current. From these 
observations the theory of electrolytic dissociation, or the 
ionization theory , was developed. 



THEORIES OF AQUEOUS SOLUTIONS 


9 


When substances like hydrogen chloride^ HC 1 , sodium 
hydroxide, NaOH, or sodium sulphate, Na 2 S04, are dissolved 
in water, the molecules break up or dissociate into charged 
atoms or radicals called ions. Positive and negative ions are 
produced, the number of positive charges being equal to the 
number of negative charges and they are of the same mag¬ 
nitude, so that the whole solution is electrically neutral. The 
substance is ionized, and the ionization is greater the more 
dilute the solution. As the concentration becomes less, the 
ionization approaches a maximum, at which point the sub¬ 
stance is completely ionized. If the solution becomes more 
concentrated the ionization is decreased, so that electrolytic 
dissociation is a reversible process. 

If a current of electricity is passed between two poles 
placed in such solutions, the positively charged ions (cations) 
move toward the negative pole (cathode); the negatively 
charged ions (anions) move toward the positive pole (anode). 
The ions therefore migrate in opposite directions, and to a 
certain extent are independent of each other. In the case 
of hydrochloric acid we have the reversible dissociation: 

HC 1 ^H +1 +CH. 

If a current of electricity is passed through the solution, the 
cations (H fl ) move to the cathode, and the anions (Cl -1 ) 
move to the anode. At these poles the ions give up their 
charges and the neutral molecules H 2 and Cl 2 are formed; 
i hydrogen and chlorine gases are evolved. This is electrolysis. 

Those substances which give low molecular weights by the 
osmotic pressure method are salts, acids, and bases. All acids 
yield hydrogen ions (as above) and all bases yield hydroxyl ions: 

NaOH^±Na +1 +OH -1 . 




10 


QUALITATIVE CHEMICAL ANALYSIS 


The charges on the ions Na +1 and OH -1 are equal and oppo¬ 
site as above indicated, and various ions have charges in 
proportion to their valence. Thus, the copper ions Cu+ 2 in 
a solution of copper sulphate, CuS 0 4 , have twice the charge 
of the hydrogen ion, etc. Twice the amount of electricity is 
required to liberate a gram atom of copper from a solution of 
copper sulphate as is needed for the liberation of one gram 
atom of hydrogen from hydrochloric acid or any other acid. 
Valence, therefore, may be considered to be the power an 
element possesses of assuming a certain number of electrical 
charges which may be either positive or negative. 

96,530 coulombs of electricity will liberate one gram atom of hydro¬ 
gen (1.008 grams) from a solution of HC 1 , but only ^ a gram atom of 
copper (63.57/2 or 31.79 grams) from a solution of copper sulphate. 
CuS 0 4 . Copper in copper sulphate, therefore, has twice the electrical 
charge in solution as hydrogen in hydrochloric acid, and its valence 
is twice as great. 

The reactions most frequently involved in qualitative 
analysis are those between ions—the formation of neutral 
substances by the mutual loss of opposite charges. Thus 
the production of a precipitate of barium sulphate by the 
action of sodium sulphate on barium chloride may be illus¬ 
trated thus: 

BaCl 2 ^Ba+ 2 +2Cl" 1 , 

Na 2 S04^2Na+ 1 +S0 4 - 2 , 

Ba+ 2 +2 CH + 2 Na +1 +S 0 4 " 2 BaS 0 4 + 2 Na +1 + 2 CH. 

The barium and sulphate ions are discharged and form the 
neutral molecule of barium sulphate which precipitates out. 
(Not all of the barium and sulphate ions are discharged; 
see solubility-product below.) Since the reactions of acids, 
bases, and salts are the result of the interaction of their 


THEORIES OF AQUEOUS SOLUTIONS 


11 


ions, the activity of such electrolytes will be largely deter¬ 
mined by the extent to which they are dissociated. By 
comparing the electrical conductivity (which, of course, is 
greater the more charged particles present) of a solution of 
an electrolyte with the conductivity of that same amount 
of electrolyte at infinite dilution (where ionization is com¬ 
plete), the percentage dissociation may be calculated for any 
definite concentration. It is found that many of the mineral 
acids such as HC 1 , HNO3, H2SO4, the alkali hydroxides KOH 
and NaOH, and nearly all salts are strongly dissociated, 
whereas H2S, HCN, many organic acids, NH4OH, etc., are 
but little dissociated. The limiting conductivities of the strong 
electrolytes are reached at finite concentrations, whereas the 
complete dissociation of such weak electrolytes as acetic 
acid is obtained in such dilute solutions that the conductivity 
has to be calculated by indirect methods. The table (p. 165) 
showing comparative ionization values for common acids, 
bases, and salts, should be studied. 

The concentration of the solutions are expressed in terms of nor¬ 
mality. A normal solution is one which contains in one liter an amount 
of substance equivalent to one gram atom of hydrogen. Thus a liter 
of a normal solution of hydrochloric acid contains 36.46 grams of hydro¬ 
gen chloride; a liter of one-tenth normal solution contains 3.646 grams 
of hydrogen chloride, etc. In the reaction: 

HC 1 +NaOH -* NaCl +H 2 0 , 

- one gram molecular weight of NaOH is equivalent to one gram molec¬ 
ular weight of hydrogen chloride, or to one gram atom of hydrogen. 
Hence a normal solution of NaOH should contain one gram molecular 
weight or 40.01 grams of NaOH per liter. 

The student should note that if the solution of an electrolyte is 
diluted, the percentage ionization increases , but the actual concentration 
of the ions —the number in the same volume of solution— decreases. 


12 


QUALITATIVE CHEMICAL ANALYSIS 


The conductivity of a solution is the reciprocal of the electrical 
resistance which may be experimentally determined. To obtain com¬ 
parative values, molecular conductivities are calculated. Molecular con¬ 
ductivity or conductance is the conductivity of that amount of solu¬ 
tion (placed between electrodes one centimeter apart) which contains 
one mol of the electrolyte. In the following table giving values of 
conductivity and ionization for acetic acid solutions of decreasing 
concentration, v is the volume containing one mol of acetic acid, the 
normality signifies the strength of the solution as defined above, fXv is 
the molecular conductivity, and a is the percentage ionization. The 
limiting conductivity , of acetic acid may not be directly determined 
as stated above, but has been found to be 361 by methods which 
need not be here described. 

MOLECULAR CONDUCTIVITY AND PERCENTAGE IONIZATION OF 
ACETIC ACID FOR DIFFERENT DILUTIONS. TEMP., 25 ° C. 


Moo =361. 


V 

Normality 

M» 

at 

8 

O.13 

4-342 

1.20 

32 

O.031 

8.70 

2.41 

128 

O .0078 

17.11 

4-74 

512 

O.OO19 

33-24 

9.21 

1024 

O.OOIO 

45-87 

12.71 

2048 

0.0005 

63.00 

17-45 


It may be seen from the table that when one gram molecule of acetic 
acid is dissolved in 2048 liters of water, the molecular conductivity and 
the dissociation are far from the maximum. 

The method of calculating percentage ionization may be illustrated 
by the value for a solution of 0.13 normality: 

H V 4-342 

a = 100 X— = 100 — = 1.20. 

Moo 361 

The following table presents data for the highly ionized electrolyte 
hydrochloric acid: 










THEORIES OF AQUEOUS SOLUTIONS 


13 


MOLECULAR CONDUCTIVITY AND PERCENTAGE IONIZATION OF 
HYDROCHLORIC ACID FOR DIFFERENT DILUTIONS. TEMP., 
2 S ° C. 


V 

Normality 


O' 

128 

O.OO78 

384 -9 

99.42 

512 

O.OOI9 

392.0 

100.0 

The maximum 

conductivity of 

hydrochloric acid 

is reached at a 


concentration such that the solution is about 0.0019 normal. At this 
concentration hydrochloric acid is completely ionized, whereas acetic 
acid is only 9.21 per cent ionized. 

9. Reaction Velocity and Law of Mass Action. There is 
another factor besides that of simple ionization which is of 
great importance in the study of chemical reactions. When 
phosphorus pentachloride is vaporized, as we have seen, it 
dissociates into phosphorus trichloride and' chlorine. The 
reaction is reversible: 

PCI5 PCI3+CI2, 

and the higher the temperature the more complete the 
dissociation. At any definite temperature it has been found 
that the rate of reaction in such cases is proportional to the 
concentration of the reacting substances. In general this law 
—the law of mass action —requires such a relation to hold 
whether the reacting substances are molecules or ions. In 
the example given above, if we start with a definite amount 
of chlorine and phosphorus trichloride they will react with 
a velocity which decreases as their concentration becomes 
lower. As the phosphorus pentachloride increases in amount 
the reverse action of its dissociation is accelerated according 
to the above experimentally proved law. At a perfectly 










14 


QUALITATIVE CHEMICAL ANALYSIS 


definite concentration the rates of the opposing reactions 
become equal; here just as much PCls is decomposing into 
PCI3 and CI2 as is being formed by their combination in the 
same time, and we have a state of dynamic equilibrium. At 
this point the ratio of the product of the concentrations of 
the reacting substances on one side of the chemical equation 
to the product of the concentrations of those on the other 
side is a constant, which follows from the law of mass action. 
In this case: 

[PCl 3 ]X[Cl 2 ]_ r 

[PC1 5 ] 

where the bracketed formulas signify the number of gram 
molecules, and K is a constant. It can be seen that if a 
system like this is in equilibrium and we add more of one 
reacting substance, say chlorine, then the equilibrium will be 
displaced; more PCI5 will be formed in order that the above 
ratio should remain a constant, and the dissociation of the 
PCI5 is thus suppressed. 

At any definite temperature, by the mass law the velocity of reac¬ 
tion is proportional to the concentration of the reacting substances. 
If we take the reversible reaction: 

A+5±=>C+A 

and start with a mixture which contains only A and B, these two 
substances will react to form C and D with a velocity depending on 
their concentration (expressed in mols): 

Velocity 1 = KiXConc. A X Conc . B , 

where A, is a constant dependent on the nature of the substances. It 
is the characteristic velocity constant, and is equal to the velocity of 
the reaction when the concentration of the reacting substances is unity. 

As the reaction proceeds, the amounts of A and B, consequently 
also Velocity h will decrease. 



THEORIES OF AQUEOUS SOLUTIONS 


15 


Since C and D as they are formed tend to unite to form A and B, 
the velocity to the left will be at any time: 

V elocity* = K 2 X Conc.c X Conc.D . 

This velocity will be small at first, but as C and D increase in amount 
the velocity proportionally increases. At a certain point the velocity 
with which A and B interact will be equal to the velocity with which 
C and D interact. Just as much of C and D will be formed as decom¬ 
pose into A and B. The actual concentrations of the four substances 
will not change, although we conceive of this point as one where the 
opposing reactions are still going on, but in effect neutralize each other. 
It is a dynamic equilibrium. 

Combining the above two equations, at equilibrium (where the veloci¬ 
ties are equal): 


and 


Cone .a xConc.BXKi = Conc.cXConc.nXK 2j 

Conc.c X Conc.D 
Conc.AXConc.B K 2 


K being a new constant—the so-called equilibrium constant. No 
matter whether we take C and D alone and allow them to interact, 
or mix any quantities of A, B, C and D, at equilibrium the four sub¬ 
stances will be present in such concentration that the same numerical 
value will be obtained for the constant K. 

A good illustration of the application of the above equation is the 
condition of equilibrium of dissociated hydrogen iodide vapor: 

2HI<=±H 2 +I 2 . 


The velocity of dissociation (reaction to the right) may be expressed 
thus: 


V elocityx = KiX Concern. 


(Where two molecules of a substance appear in a reaction, the con¬ 
centration is squared, where three appear it is cubed, etc.; in this 
case the hydrogen iodide has the same effect as if it were acting twice, 
or as two substances, thus: 


HI-fHI^H 2 +I 2 ). 



16 


QUALITATIVE CHEMICAL ANALYSIS 


The velocity with which hydrogen and iodine combine will be: 


At equilibrium: 


V elocityi = K 2 X Conc.H 2 X Conc.j 2 . 


Conc.H 2 XConc.i 2 
Cone . 2 hi 


=K. 


K is the equilibrium, or in this case, the dissociation constant, and 
will be dependent on the temperature, but not on the initial amounts 
of reacting substances. 

K for the above reaction varies with the temperature as follows: 


Temp.280° 3 20 0 360° 400° 448° 500° 

K .0.012 0.013 0.015 0.017 0.020 0.025 


The constancy of K for any one temperature is shown in the fol¬ 
lowing table. Known quantities of hydrogen were allowed to act on 
iodine at 448° and at equilibrium the amount of hydrogen iodide formed 
tvas determined. In the table the concentrations of hydrogen and 
iodine are the original amounts used, the concentration of HI is the 
amount formed. 


K for 2HI H2+I2 at 448°. 


Cone. H2 in c.c. 

Cone. I2 in c.c. 

C.c. HI formed 

K 

20.60 

14-45 

25.72 

O.OI9 

14.47 

38.93 

27.64 

0.021 

8.07 

9.27 

13-34 

0.020 


In the equilibrium equation: 

Conc.H 0 XConc.i 

_ i _ -=K 

Cone . 2 hi 

the concentrations are understood to be expressed in mols. But the 
same number of molecules are contained in equal volumes of all gases 
under the same conditions of pressure and temperature, so that we may 
substitute volumes (reduced to o° and 760 mm.) directly in the equilib- 












THEORIES OF AQUEOUS SOLUTIONS 


17 


rium equation instead of concentrations expressed in mols. At equilib¬ 
rium the concentration of the hydrogen in the first experiment given 
in the table is 20.60—25.72/2 = 7.74 c.c., since the volume of hydrogen 
disappearing in the reaction is only one-half that of the hydrogen iodide 
formed. Likewise the volume of iodine vapor at equilibrium is 14.45 
— 25.72/2 = 1.59 c.c. Therefore 7.74X1.59/25.72 2 = A=o.oi9. 

The law of mass action may be applied to many revers¬ 
ible reactions; especially important to the qualitative analyst 
are those which take place in solution. 

Ionization is a reversible process and here the law may 
be applied. For acetic acid: 


HC2H3O2 H+i + CsHsCV 1 . 

At a definite temperature then, the equilibrium may be ex¬ 
pressed thus: 

[H +1 ] X [C2H3O2- 1 ] _ r^. 

[HC2H3O2] 

the ratio of undissociated acid to the concentration of its ions 
is constant and independent of the dilution. This so-called 
ionization constant is a measure of the activity of the acid. 
With an acid more highly ionized, the constant would be 
larger. With the base ammonium hydroxide: 

[NH 4 +1 ]X[ 0 H~ 1 ] _ 

[NH4OH] 

and the constant is a measure of the activity of the 
base. 

The validity of the mass law as applied to certain electrolytes may 
be shown by acetic acid, whose ionization constant for different con¬ 
centrations of acid is given in the following table: 




18 


QUALITATIVE CHEMICAL ANALYSIS 


V 

Normality 

a 

K 

8 

O.13 

I . 20 

0.000018 

32 

O.O31 

2.41 

0.000018 

128 

O.OO78 

4-74 

0.000018 

512 

O.OOI9 

9.21 

0.000018 

1024 

O.OOIO 

12.71 

0.000018 

2048 

0.0005 

17-45 

0.000018 


The constant may be calculated as follows: For a solution of 0.13 
normality, which contains one mol of acetic acid in 8 liters,""since it 
is 1.20 per cent dissociated the concentration of both H +1 and CaHsOr 1 
(equivalents per liter) will be 1X0.012 = 0.0015. This leaves 0.0015 
= 0.1235 equivalents of undissociated HC 2 H 3 0 2 . Substituting in 
H +i X C 2 H 3 0 2 _1 /HC 2 H 3 0 2 = K, 0.001 5 2 /o. i 235 = K = 0.000018. 

It is a striking fact that the mass law does not hold for 
strong electrolytes. The ionization constants of strong acids, 
bases, and of most salts are not constants at all, but vary 
with the dilution. While this must be borne in mind by the 
student, it does not lessen the value of the application of 
the mass law to systems for which it is valid. The student 
should consult references 1 for data on the ionization con¬ 
stants of various acids and bases. 

10. Common-ion Effect. If sodium acetate is added to 
a solution of acetic acid, the concentration of the acetate 
ions is increased, and for the equilibrium ratio 

[H+ 1 ]X[C 2 H 30 2 - 1 ] = 

[HC 2 H 3 0 2 ] A ’ 

to remain constant, more undissociated acetic acid must 
form with a disappearance of hydrogen ions. This is found 
experimentally to be the case, and the acidity of the solu- 
1 See Stieglitz. Elements of Qualitative Analysis, I, pp. 104-106. 










THEORIES OF AQUEOUS SOLUTIONS 


19 


tion is lowered. In a similar way, the addition of ammonium 
chloride to a solution of ammonium hydroxide decreases the 
basicity of the latter. By adding a common ion the ioniza¬ 
tion of an electrolyte is reduced. This is the common-ion 
effect. (Other factors such as formation of complex ions 
may complicate this phenomenon. Examples will be encoun¬ 
tered later in the course of study.) 

If a liter of solution is normal with respect to both acetic acid and 
sodium acetate—that is, if it contains one mol of each—the solution 
is less acid than a normal solution of pure acetic acid, which is 0.42 
per cent ionized at room temperature. Since a normal solution of 
sodium acetate is 53 per cent ionized, the total amount of acetate 
ions in the mixture, expressed in gram equivalents, will be 0.53+0.0042 
= 0.5342. Since the equilibrium ratio, [H 41 ] x[C2H30£ -I J/[HC 2 H 3 0.2] 
= 0.000018 (§9) should hold true, no matter what salts are added, the 
amount of H +1 present will be [H +1 ] = 0.000018 X[HC 2 H30 2 ]/[C 2 H302 _1 ] 
= 0.000018 X0.9958/0.5342 = 0.000034. The H +1 ions are thus 0.000034/ 
0.0042 = 0.008 times .their original value. The ionization of the acetic 
acid has been reduced from 0.42 per cent to 0.0034 per cent. Experi¬ 
mental facts confirm the theory. 

In the above deduction, the amount of acetate ions from the acid, 
and the amount of undissociated acid are assumed constant and un¬ 
changed when the salt is added. However, if hydrogen ions disappear, 
they do so by forming undissociated acetic acid: 

H+i+CAOr 1 <=± HC 2 H 3 0 2 . 

These effects are so small compared with the reduction in hydrogen 
ions that they may be disregarded in the calculation. It should also 
be noted that the ionization of sodium acetate would be lowered by the 
presence of the common ion in acetic acid, but this action is also in¬ 
appreciable because of the large ionization of the salt. 

11. Solubility-product Principle. In the consideration of 
substances suspended in water, it is assumed that none is 
completely insoluble. Magnesium hydroxide is quite insolu¬ 
ble in water, but not completely so. Some of the hydroxide 


20 


QUALITATIVE CHEMICAL ANALYSIS 


dissolves, yielding undissociated magnesium hydroxide in 
solution, which dissociates into magnesium and hydroxyl 
ions: 

Mg(OH) 2 <=* Mg +2 + 20H -1 , 

and 

[Mg +2 ] X [OH -1 ] 2 _ ^ ^ 

[Mg(OH) 2 ] . V 

But the saturated solution is in contact with solid Mg(OH) 2 
and there is a second equilibrium in which solid Mg(OH) 2 
is dissolving at the same rate at which it is being precipitated. 
We have then: 


and 


Mg(OH) 2 solid Mg(OH) 2 Dissolved ^ Mg +2 +2 OH -1 , 
[Mg(OH) 2 Dissolved] _ -rr 

[Mg](OH) 2 bom] 1 . 


( 2 ) 


But the concentration of the solid is a constant; that is, it 
is experimentally verified that the solubility of a substance 
is independent of the amount of solid in contact with the 
solution. Therefore 

lMg(OJT) 2 Dissolved] = A 2 .(3) 

and from our original equation (1), 

[Mg+ 2 ] X [OH -1 ] 2 = K3. 


This is the solubility product. The product of the ions of a 
substance in solution which is in equilibrium with the solid 
substance yielding these ions is constant. 

Precipitation may be explained as follows. When NH 4 OH is added 
to MgCl 2 , the concentration of the hydroxyl ions and of the magnesium 
ions is sufficient to exceed the above solubility product, and therefore 







THEORIES OF AQUEOUS SOLUTIONS 21 

they must unite to form Mg (OH) 2. The solution becomes saturated 
with this and it precipitates out. 

MgCL +2NH4OH—> Mg(OH) 2 +2NH 4 Cl 

or 

Mg +2 + 2 OH- 1 -» Mg(OH) 2 . 

Now if a sufficient amount of an ammonium salt is added to NH 4 OH, 
the latter will not precipitate Mg(OH) 2 from solutions of MgCl 2 . The 
addition of ammonium ions has decreased the hydroxyl ions of the 
base by the common ion effect, and there are insufficient hydroxyl 
ions for the solubility product of Mg(OH) 2 to be exceeded. 

If HC1 is added to Mg(OH) 2 the hydroxide dissolves: 

Mg(OH) 2 +2HCl—* MgCl 2 + 2H 2 0. 

The H +1 ions take up the OH -1 ions to form water, and the OH -1 
ions are reduced so far that the equilibrium as expressed by the solu¬ 
bility product cannot be maintained, and the Mg(OH) 2 passes com¬ 
pletely into solution. 

Many applications of the solubility product principle may be made 
in qualitative analysis and some of them will be discussed later. One 
of the most important consequences of the principle (where no other 
factors are involved) is that the solubility of a salt is decreased by 
adding another electrolyte with a common ion. Thus the solubility of 
the difficultly soluble salt silver chloride, AgCl, is 1.25 Xio -5 mols per 
liter at 25 0 . Therefore the solubility product will be 

[Ag +1 ] X[C1 ~ 1 ] = A'=[i. 2S Xio~ 5 ] 2 = i.56Xio -10 . 

If to a liter of the saturated solution, five times as much sodium chloride 
is added as there is AgCl in solution, the concentration of the CL 1 
ions will have been increased to about 5 times their original amount; 
for the solubility product to remain constant, the Ag +1 ions—con¬ 
sequently the AgCl in solution—will be reduced to approximately one- 
fifth of the original concentration. (The concentration of the undis¬ 
sociated salt is inappreciable compared with that of the ions.) The 
silver chloride is thus made more insoluble or, in other words, the 
precipitation is more complete . 

12. Hydrolysis. We have finally to consider at this 
point the action of water on salts as intensified in certain 


22 


QUALITATIVE CHEMICAL ANALYSIS 


cases. Water is dissociated to a slight extent into hydrogen 
and hydroxyl ions: 

H 2 0 ^±H+ 1 H- 0 H- 1 , 


and 


[H +1 ]x[OH~ 1 ] _ „ 

[H 2 0 ] 


If a salt of a weak acid, for example KCN, is dissolved in 
water, the solution is found to be strongly alkaline, showing 
that hydroxyl ions are present. This is due to the hydrolysis 
of the potassium cyanide— the breaking up of the salt by 
water into base and acid. KCN will dissociate thus: 


KCN^K+i + CN- 1 .. 

and the ions of water will combine with these potassium and 
cyanide ions: 

K +1 +OH -1 <=± KOH, 

CN -1 +H +1 <=± HCN, 
or 

KCN+H2O <=> KOH+HCN. 

Hydrocyanic acid is a very weak acid, and therefore a con¬ 
siderable number of CN -1 ions will unite with the H +1 ions 
of the water in the formation of undissociated HCN. As 
the H+ 1 ions are taken up, the water will continue to dis¬ 
sociate in an attempt to satisfy its equilibrium requirements, 
and the hydrolysis will proceed. Potassium hydroxide is 
practically completely ionized so that there is little union 
between the K +1 ions and the OH” 1 ions to form undis¬ 
sociated KOH. The increase in OH -1 ions will limit the 
production of HCN by suppressing the H +1 ions of the water 
by the common ion effect. At equilibrium, there will be 
an excess of OH -1 ions, and the solution will react alkaline. 



THEORIES OF AQUEOUS SOLUTIONS 


23 


The strength of acids varies as their ionization constants (§ 9 ), and 
when two acids are added to a base, they compete for it in such a 
way that the base is .distributed between the acids in the same ratio 
as their constants. Hydrolysis of the salt of a weak acid may be 
considered in this light since water is a very weak acid. The dissocia¬ 
tion constant for water, 

\H+']X[OHT l ]=K, 

the concentration of the undissociated water is considered constant 
since it is so large compared with the concentration of its ions) has 
been found to be about 1.2 Xio -14 at 25 0 . Hence 

[H +1 ] = [OH -1 ] = V1.2X io -14 = 1.1 X io -7 ; 

or there are 1.1X10 -7 gram equivalents of H +1 ions in one liter of 
water. We have, then, the weak acid HCN competing with the weak 
acid HOH for the base KOH, when KCN is dissolved in water. 

From the equation 

KCN +H 2 0 <=> KOH +HCN, 

we have the general equation following directly from the mass law: 

[Base] X [Acid] 

[Unhydrolyzed Salt] h * 

where Kh is the hydrolysis constant, independent of the concentrations 
of the substances at equilibrium. A 0.095-normal solution of KCN 
is hyd.olyzed i.12 per cent at 25 0 . Therefore [Base] = [Acid] = 0.095 
Xo.oi 2 = 0.001064, and [Unhydrolyzed Salt] = 0.095—0.001064 = 0.094. 
Hence K h = (o.coio64) 2 /o.o94= 1.2Xio -5 . 

To derive the relation between the hydrolysis constant and those 
for HCN and water, we divide the equation expressing the equilibrium 
for water: 

[H +1 ]X[OH -1 ] = Aw a ter, 

by that for HCN : 

[H +1 ] x[CN -1 ] ^ 

[HCN] Kkcn ’ 

and obtain 

[HCN] x[OH~ ^] _ K water 
[CN -1 ] Ahcn 







QUALITATIVE CHEMICAL ANALYSIS 


But [HCN] x[OH _1 ]/[CN -1 ]=A:a from the general equation of hydrolysis 
given above, and 


The hydrolysis constant is measured by the ratio of the ionization 
constants for water and the acid, and the hydrolysis will be greater 
the weaker the acid. 

Salts of strong acids with weak bases will likewise. hydro¬ 
lyze considerably, since the cations of the salt and the OH -1 
ions of the water will tend to unite to form the slightly 
dissociated base. Such solutions will react acid, and the 
hydrolysis will be limited by the accumulation of H +1 ions 
from the acid formed. 

Here we have the weak base water, competing with a weak base 
such as ammonium hydroxide formed by breaking up ammonium 
chloride which we may take as an example, for the acid HC1: 

NH 4 C 1 +H 2 0 <^ NH4OH+HCI. 


It may be shown that the hydrolysis constant equals the ratio of 
the dissociation constant for water to that of ammonium hydroxide: 


K h = 


K water 

Anh 4 oh 


Salts of weak bases with weak acids will be very largely 
hydrolyzed for obvious reasons. Further, when the base 
or acid formed is insoluble, the hydrolysis may be prac¬ 
tically complete. Thus when ammonium sulphide is added 
to an aluminium salt, aluminium sulphide is not produced, 
but aluminium hydroxide, since aluminium sulphide is the 
salt of a weak, insoluble base and a weak, insoluble acid. 
It therefore is hydrolyzed: 

AI 2 S 3 + 6 HOH 2A1(0H) 3 +3H 2 S. 




THEORIES OF AQUEOUS SOLUTIONS 


25 


The dissociation of water increases with rise in tempera¬ 
ture. Heating and employing an excess of water favor 
hydrolysis. Thus if a dilute solution of ferric acetate is 
boiled (preferably with the addition of a salt of acetic acid 
to reduce the ionization of the acetic acid formed in the 
hydrolysis) a basic ferric acetate is precipitated: 

Fe(C2H 3 02)3+2H0H^Fe(0H) 2 (C2H302) + 2HC2H^02. 

13. The Electron Theory of Matter and Its Bearing on Electro¬ 
lytic Dissociation. A recent theory of matter is that the atoms consist 
of smaller particles of matter or corpuscles, and that these corpuscles are 
negative electricity existing in a sphere of positive electrification; to 
these corpuscles is given the name electrons to signify their nature. The 
electron is about one-thousandth of the mass of the hydrogen atom and 
has a charge equal to the charge of any univalent ion. In an isolated 
atom the sum -of the charges of the electrons equals the total positive 
electricity and, therefore, the atom is neutral. The atoms differ from 
each other in the number of electrons and probably in the arrangement 
in the atom. 

Now the elements appear to possess varying power to give off or 
take on one or more electrons. We may consider that in compounds 
the elements are held together by opposite charges gained thus: in 
sodium chloride, NaCl, the sodium atom has given up one electron to 
the chlorine atom, which latter will thereby have an excess of negative 
electricity and will be accordingly charged with a charge equal to that 
of one electron. By losing one negative charge the sodium atom has 
an excess of positive electricity of equal amount. Thus: 

Na -t Cl +e =Na +1 Cl -1 , 

where e is the symbol for one electron. The formula Na +t Cl -1 signifies 
there is one positive charge on the sodium atom and one negative charge 
on the chlorine atom. In ferric chloride we have 


Fe-*‘Cl3 +t =Fe +3 Cli“ I . 

Therefore valence is the power of elements to give off or to assume a 
definite number of electrons. 


26 


QUALITATIVE CHEMICAL ANALYSIS 


When electrolytes are dissolved in water, they are dissociated as 
we have seen. Under the electron theory, the dissociation consists 
in separating the charged atoms from each other by the influence of 
water, so that these charged atoms may act independently as ions: 

Fe +3 C 1 3 -1 Fe +3 +3CI -3 . 


GENERAL INSTRUCTIONS 

14. Make two 500 c.c. wash-bottles, one for hot water 
and one for cold water (distilled water is used in all opera¬ 
tions). 

Make several stirring rods of convenient length by round¬ 
ing off the ends of glass rods. 

The strength of desk, side-shelf, and special reagents is 
given on p. 155, in terms of normality, and frequent 
reference to these figures should be made. A normal solu¬ 
tion has been defined in (8). Solutions of equal normality 
are equivalent; thus 1 c.c. of 6-N (six-normal) hydrochloric 
acid will neutralize 1 c.c. of 6-N sodium hydroxide solution. 
Likewise 1 c.c. of normal hydrochloric acid will precipitate 
qualitatively (not quantitatively on account of the reversi¬ 
bility of reactions, etc.) the silver as chloride from 1 c.c. 
of normal silver nitrate solution. 

By knowing the concentrations of reagents and test solu¬ 
tions, the student may learn the sensitiveness of reactions. 
If a reagent gives with a certain test solution a characteristic 
test, by dilution of the test solution to a known volume, the 
sensitiveness of the reaction at this known dilution may be 
determined by the result observed. • 

Reagents as mentioned below are ordinarily the dilute 
solutions; concentrated solutions are specifically - called for. 
Thus by HC 1 is meant the dilute desk-shelf reagent, 6-N 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 27 

hydrochloric acid; by cone. HC 1 is meant the concentrated 
13-N hydrochloric acid. Abbreviations for precipitate and 
residue are ppt. and res. respectively. 

For the test reactions use test-tubes, diluting the strong test 
solutions (usually the nitrates, N/2) 3 or 4 times. To 2-3 c.c. of 
the diluted solution add the reagent drop by drop unless other 
directions are given. If a precipitate forms, let it settle, and 
then in the clearer supernatant liquid a few drops more of the 
reagent will indicate whether precipitation is complete. 

Write in the notebook the equations for every reaction 
together with the observations in each experiment—color and 
character of the precipitates formed, solution phenomena, etc. 
For brevity, statements of certain results of an experiment are 
frequently substituted for specific directions. Thus, in Exp. 6 
(15), treat mercurous nitrate solution with potassium hydrox¬ 
ide solution, note the color of the precipitate, and write the 
equation for the reaction involved and also for the decomposi¬ 
tion of mercurous oxide. Answer any questions. 

REACTIONS OF THE BASE-FORMING CONSTITUENTS, AND 
BASIC ANALYSIS 

Group I. The Silver Group. Hg +1 , Pb+ 2 , Ag +1 . 

Reactions 

MERCURY 

15. Mercury may have a valence of one 1 or two in its 
compounds, and has two oxides, black mercurous oxide/ 

1 The question of the structure and valence of the mercurous ion is still 
an open one. The mercurous ion is often assumed to be divalent with the 
structure —Hg—Hg—. Analytical facts, however, may be explained by assum¬ 
ing the ion Hg +1 . 


28 


QUALITATIVE CHEMICAL ANALYSIS 


Hg 2 0, and yellow or red mercuric oxide, HgO. The hydrox¬ 
ides are unstable and break up into the oxides and water 
when attempts are made to prepare them. Mercurous and 
mercuric ions, corresponding to the two classes of mercury 
salts, are colorless. 


MERCUROUS SALTS 

1. Heat some calomel, Hg 2 Cl 2 , in a hard glass tube 
closed at one end (closed tube test). 

Mercury compounds sublime unchanged when heated, or decompose 
with the formation of metallic mercury. 

2 . Heat a dry mixture of Hg 2 Cl 2 and Na 2 C0 3 in a closed 
tube. 

All mercury compounds heated with sodium carbonate in a limited 
supply of air yield metallic mercury. 

3 . Treat about 0.5 gram dry Hg 2 (N0 3 ) 2 with 5 c.c. water, 
boil, and test the solution with litmus paper. 

Many basic salts may be formed, but in this case write the reaction 
expressing the hydrolysis on the assumption that Hg 2 ( 0 H)(N 0 3 ) is 
formed. 

Indicators are weak acids or bases whose anions or cations, respect¬ 
ively, are of different color from the undissociated molecule. If a 
base is added to a solution containing an acid indicator, e.g., phenol- 
phthalein, whose undissociated molecule is practically colorless a salt 
wiH form which will dissociate and yield the anion of the indicator 
which m this case is red to pink, and the solution wiU be of this color. 
Changes in the structure of the molecule accompany change in color 
which only a knowledge of organic chemistry would make clear Indi¬ 
cators are more or less sensitive to H+ 1 and OH- 1 ions and the’follow¬ 
ing table shows the approximate concentration of these ions at which 
there is a pronounced color change. Concentrations are expressed in 
gram equivalents per liter. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 29 


Indicator 

+ 0 
E x 

„ 7 
+ 2 

7 - 0 

ffi X 

+ 2 
W X 

+ 2 
KX 

8a 

+ 2 
w 

T 1 
l 0 

O x 

H 

v 7 

1 0 

Q x 

T 0 
”8 

7 0 
a « 

° x 

Methyl violet. . . 
Methyl orange. . 

Congo red. 

Alizarine. 

Litmus. 

Phenolphthalein 

Blue 

Violet 

Red- 

orange 

Blue 

Orange 

Violet 

I Yellow 

Red 

Brown 

Red 

Red 

Violet" 

Blue 

No 

Color 

| Pink 

Red 

Violet 

. 




4 . Hg 2 Cl 2 is precipitated from solutions of mercurous 
salts by HC1. 

Calomel is somewhat soluble in cone. HC 1 and excess of alkali 
chlorides due probably to the formation of complex ions. 

Complex ions are those in which a certain ion has been combined 
with another elementary atom, ion, or group, whereby the new ion does 
not give the reactions of the original ion. Thus, if H 2 S is oxidized to 
H2SO4, the new sulphate ion, S 0 4 ~ 2 , will not precipitate CuS from solu¬ 
tions of copper salts as the original sulphide ion, S -2 , has the power to do. 

With excess NaCl, e.g., calomel forms complex salts and the reactions 
may be expressed as follows: 

Hg 2 Cl 2 <=>Hg+HgCl 2 

or 

2 Hg +1 ^±Hg+Hg+ 2 . 

Now if the Hg +2 ions are decreased, the equilibrium will be displaced 
according to the law of mass action so that the reaction proceeds more 
to the right. The Hg +2 ions may be taken up thus: 

HgCl 2 + 2NaCl <=> Na 2 HgCl 4 

or 

Hg +2 + 4 Cl- 1 ^(HgCl 4 )- 2 . 

The entire reaction, expressed in the non-ionic form, is 
Hg 2 Cl 2 +2NaCl Na 2 HgCl 4 +Hg. 

The complex ion formed is (HgCK) -2 . 




















30 


QUALITATIVE CHEMICAL ANALYSIS 


5. Add NH4OH to mercurous nitrate and to calomel. 
The compounds . NH 2 HgN03*Hg0 and NH 2 HgCl may be 
assumed. 

NH4OH produces usually black precipitates with mercurous salts, due 
to the formation of amino mercuric compounds (—NH 2 is the amino 
group) and metallic mercury. This is the result of the breaking up of the 
mercurous ion according to the reaction given in Exp. 4. 

6. KOH or NaOH precipitates mercurous oxide, Hg 2 0 . 
Hg_(OH)_. does not exist, at least at room temperature. 

7. Na 2 C03 gives a precipitate of Hg 2 CC>3 which rapidly 
decomposes into Hg, IlgO, and C 0 2 . 

8. H 2 S produces a black precipitate with mercurous salts. 

From the above observations on the stability of mercurous compounds, 
of what does this precipitate most probably consist? 

9. K 2 Cr04 precipitates Hg 2 Cr04. 

10. KI precipitates Hg 2 I 2 . 

In view of the behavior of calomel with excess NaCl as explained 
in Exp. 4, what would be the probable action of excess KI on Hg 2 I 2 ? 

11. SnCl 2 reduces mercurous salts to metallic mercury. 

Oxidation and Reduction. The equation for the reaction with 

calomel is 

SnCl 2 +Hg 2 Cl 2 —>SnCb+2Hg. 

Hg 2 Cl 2 , suspended in water, yields a few Hg +1 and Cl -1 ions (solu¬ 
bility product p. 19): 

Hg 2 Cl 2 2 Hg +1 + 2 C 1 -1 , 

so the reaction may be expressed as between ions: 

Sn +2 +2Cl“ 1 +2Hg +1 +2Cr 1 -^ Sn +4 + 4 Cl~ 1 + 2Hg°, 

or simply 

Sn +2 +2Hg +1 —,Sn+ 4 +2Hg°. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 31 


The only change involved is the increase in valence of the tin ions 
from 2 to 4, and the decrease in valence of the mercury from i to o. 
The positive charges of the mercury ions have been transferred to the tin 
ions. The tin has been oxidized and the mercury reduced. 

If a strip of metallic copper is placed in a solution of a mercurous 
salt, mercury is deposited on the copper and an equivalent amount 
of the latter goes into solution. 

Cu+Hg 2 (N 0 3 ) 2 -> Cu(N 0 3 ) 2 + 2Hg 

or 

Cu°+2Hg +1 -> Cu +2 +2Hg°. 

The charge on the copper has been raised from o to 2, and that of the 
mercury lowered from i to o. The copper has been oxidized and the 
mercury reduced. 

Mercury dissolves in hydriodic acid with the evolution of hydrogen, 

2Hg-f 2 HI —> Hg 2 I 2 +H 2 . 

Applying the ionic theory: 

2Hg°-f2H +1 + 2l 1 —> 2Hg +1 +2l l + H 2 . 

But Hg 2 I 2 is insoluble in not too great an excess of HI, and the mer¬ 
curous and iodine ions combine to form the neutral molecule of mer¬ 
curous iodide which precipitates out. 

2Hg+ 1 +2l- 1 ->Hg 2 I 2 . 


If we consider the atoms in the neutral molecule Hg 2 I 2 held together 
by opposite charges, the entire action may be written as follows: 

2 Hg«+ 2 H + 1 + 2 r 1 _»H g 2 + 1 I 2 - 1 +H 2 . 

The mercury has gained two positive charges, and the hydrogen has 
lost two. The former has been oxidized and the latter reduced. 

Oxidation results in the gain of positive charges or the loss of nega¬ 
tive charges by atoms or ions; reduction results in the gain of negative 
charges or the loss of positive charges by atoms or ions. 

These two processes are simultaneous, and whenever there is an 
oxidation of a substance taking place, there must necessarily be a reduc¬ 
tion of another component of the reacting system. A loss of positive 


32 


QUALITATIVE CHEMICAL ANALYSIS 


charges must occur along with the loss of an equal number of negative 
charges, so that the sum total of the charges is zero. 

The electron conception of valence may be readily applied to the 
above cases. The assumption of a positive charge means the loss of 
one electron, and the assumption of a negative charge means the gain 
of one electron. 

The equation 

Sn + 2 +2Hg +1 -4 Sn + 4 +2Hg° 

may be written, 

Sn~ 2 e +2Hg~ l€ —►Sn~ 4 e +2Hg. 

The tin has given up two electrons to the mercury which becomes 
neutral. 

The reducing agent suffers a loss of electrons; the oxidizing agent 
gains electrons. 

Typical oxidizing and reducing agents used frequently in qualitative 
analysis will be discussed as they are met with. 

LEAD 

16. Lead has five oxides: black lead suboxide.. Pb 2 0 , 
yellow lead monoxide, PbO, yellow lead sesquioxide, Pb 2 0 3 , 
red lead, Pb 3 0 4 , and brown lead dioxide, Pb 0 2 . The usual 
valence of lead in its compounds is two or four. The weak 
base Pb(OH) 2 derived from PbO is stable. The stable lead 
salts are derived from this base and yield the colorless plumbic 
ion Pb +2 . The acid character of lead hydroxide, as evi¬ 
denced by the existence of plumbic ions Pb 0 2 “ 2 , will be 
discussed below. The student should consult other texts to 
recall the properties and behavior of salts and derivatives 
of the higher oxides. 

LEAD SALTS 

i. Mix any lead compound with dry sodium carbonate, 
place in a cavity in a piece of charcoal, and heat before the 
reducing flame of the blowpipe. Note the character of the 
metallic button and the color of the incrustation. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 33 

The reactions may be illustrated by the reduction of lead chloride: 

PbCl 2 -|-Na 2 C 0 3 —* PbC0 3 +2NaCl, 

PbC 0 3 —> PbO-f C 0 2 , 

PbO+C —> CO+Pb (button). 

The incrustation is PbO. In other dry tests of this kind, the student 
should describe and account for any volatile substances and incrusta¬ 
tions formed. 

The fact that the reaction 

PbCl 2 +Na 2 C 0 3 -> PbC0 3 +2NaCl 

takes place in the absence of water may nevertheless be attributed to 
an ionic interchange. It has been shown that certain salts in the 
fused condition are good conductors of electricity and even show some 
conducting power when heated to a temperature which is still below 
the melting-point. Such heated salts, therefore, have charged particles, 
and their reactions may be assumed as analogous to the reactions 
between dissolved electrolytes. 

On the other hand there have been a considerable number of reac¬ 
tions discovered in which no ionizing medium is present or any dis¬ 
sociation is observed. It seems that the ionic hypothesis cannot be 
extended to such cases. 

2. HC 1 precipitates PbCk from cold solutions of lead salts 
if these solutions are fairly concentrated. Hot water readily 
dissolves the precipitate. 

As lead chloride is not very insoluble in water, by the solubility 
product principle there are relatively considerable chlorine ions in 
solution: 

PbCl 2 ^Pb +2 +2Cl- 1 , 

and [Pb +2 ] X[C 1 _1 ] 2 =const. A moderate excess of HC 1 decreases the 
solubility of the salt by the common ion effect, as can be predicted by 
the requirements of the above equilibrium. But excess of cone. HC 1 
dissolves PbCl 2 on account of complex salt formation: 


2 HCl-f-PbCl 2 ^±H 2 PbCl 4 . 


3 4 


QUALITATIVE CHEMICAL ANALYSIS 


Test of Sensitiveness of a Reaction. To i c.c. Pb(NOs)2 
add i drop HC 1 and observe result. Dilute i c.c. Pb(NOs)2 
to io c.c., and to i c.c. of this diluted solution add i drop 
HC 1 and again observe. The weight of the lead in the two 
solutions may be calculated from the normality of the stock 
solution, which is N/2. 

The above reaction is not very sensitive and therefore would not 
serve as a good test for the presence of lead. The sensitiveness of 
the reactions of lead salts with H 2 S 0 4 and K 2 Cr 0 4 (7 below) should be 
tested, the solution of lead salt being diluted until no test is obtained. 

3. Treat Pb(NOs)2 with NH 4 OH. Pb(OH)2 is formed 
and is insoluble in excess of the reagent. 

4. Add NaOH gradually to Pb(N03)2, note the precipi¬ 
tation of Pb(OH)2, and its solubility in excess NaOH. 

In the study of the behavior of all compounds, the periodic system 
is particularly valuable, and frequent reference to it should be made 
in order to understand certain reactions and the analogous and differen¬ 
tial behavior of the elements. Lead is in the fourth group of the 
periodic system, and the basic properties of the elements increase accord¬ 
ing to the following series: Si<Ge<Sn<Pb. The acid properties 
decrease of course thus: Si>Ge>Sn>Pb. As a matter of fact, the 
behavior of certain lead compounds indicates that there is still some 
tendency for the lead to act as an acid element. This is particularly 
noticeable in its hydroxide formed thus: 

Pb(N0 3 ) 2 +2Na0H <=> Pb( 0 H) 2 + 2 NaN 0 3 . 

Lead hydroxide is basic to a certain extent, and must yield hydroxyl ions: 

Pb(OH) 2 <=± Pb +2 + 20 H- J , 

since it readily dissolves in acid: 

P b (OH) 2 +2HN 0 3 <=± Pb(N 0 3 ) 2 + 2 H 2 0 . 

But lead hydroxide is soluble in excess of NaOH, and if lead has any 
acid-forming tendency, we would have the following equilibrium: 

Pb 0 2 H 2 <r± 2 H +1 +Pb 0 r 2 , 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 35 


and lead hydroxide would be capable of acting as an acid. NaOH is 
a strong base with many hydroxyl ions which would suppress the basic 
dissociation of lead hydroxide by the common ion effect, and further 
take up the hydrogen ions with salt formation. As the H + ions are 
removed, more and more lead hydroxide would dissociate as an acid 
in order to satisfy the requirements of the equilibrium of the acid 
dissociation as expressed in the last equation. Therefore in the pres¬ 
ence of an excess of NaOH, lead hydroxide would act as an acid, 
plumbic acid: 

* H 2 Pb 0 2 + 2 NaOH <=> Na 2 Pb 0 2 + 2H0O. 

Substances like lead hydroxide, which may yield both hydrogen and 
hydroxyl ions in solution, are called amphoteric electrolytes. 

Lead hydroxide, as was observed, does not dissolve appreciably in 
NH4OH. Ammonium hydroxide is a weak base, giving far less hydroxyl 
•ions than NaOH, and would suppress the basic dissociation of Pb(OH) 2 
to a correspondingly less extent. Also ammonium plumbite formed by 
the reaction 

H 2 Pb 0 2 + 2 NH 4 0 H^ (NH 4 ) 2 Pb 0 2 + 2 H 2 0 

would be the salt of a weak base and a very weak acid. It would 
have a great tendency to hydrolyze—in other words, very little am¬ 
monium plumbite could be formed by the above reaction. Sodium 
plumbite is the salt of a strong base and a weak acid and would hydro¬ 
lyze to a less extent. However, such salts react basic when dissolved 
in water. 

Since Pb(OH) 2 is both basic and acidic, what would tend to prevent 
the formation of lead plumbite according to the reaction: 

Pb( 0 H) 2 +H 2 Pb 0 2 PbPb 0 2 + 2 H 2 0 ? 

5. Add Na 2 C 0 3 to Pb(N0 3 )2. A basic salt is precipi¬ 
tated. 

6. H 2 S precipitates PbS, insoluble in dilute HC 1 , but 
soluble in dilute HN 0 3 . (Cf. Copper.) When H 2 S is passed 
into a solution of lead chloride, a red precipitate of lead sul- 
phochloride, PbS-PbCl 2 , is first observed. A larger number 
of sulpho salts of bivalent mercury are known. 


36 


QUALITATIVE CHEMICAL ANALYSIS 


The equation for the solution of lead sulphide, 

3 PbS+ 8 HN 0 3 <=> 3 Pb(N 0 3 ) 2 + 4 H 2 0 + 2 N 0 + 3 S, 
is a result of two actions, the formation of lead nitrate, 

PbS + 2 HN 0 3 <=> Pb(N 0 3 ) 2 +H 2 S, 

and the oxidation of the H 2 S by the HN 0 3 (if the latter is fairly con¬ 
centrated), 

2 HN 0 3 + 3 H 2 S *=» 2NO +4H0O +3S. 

Such an equation as the latter may be balanced as follows: The 
nitrogen atom in HN 0 3 has five positive charges and in NO two such 
charges. There is a loss of three charges. The sulphur atom in H 2 S 
has two negative charges and none in free sulphur. There is a loss of 
two negative charges. The loss in the number of charges of opposite 
sign must be the same. (The number of electrons gained by the oxidiz¬ 
ing agent must equal the number lost by the reducing agent.) There¬ 
fore two molecules of HN 0 3 would be equivalent to three of H 2 S: 

2 HN +6 0 3 +3H 2 S -2 <=± 2N +2 0+ 4 H 2 0+ 3S 0 . 

Sometimes the following dualistic conception of such oxidations as 
the above may be of value to the student. The nitric acid may be 
considered to yield an anhydride which decomposes into nitric oxide and 
oxygen. The oxygen is taken up by the reducing agent H 2 S, and the 
sum of these actions will represent the net result: 

2 HN 0 3 -+N 2 CX+H 2 0 , 

N 2 (X -> 2NO+3O, 

3 Q+3H 2 S—> 3 S+ 3 H 2 Q,_ 

2 HN 0 3 + 3 H 2 S 2 N 0 + 4 H 2 0 + 3 S. 

7. H2SO4 precipitates the quite insoluble PbS 0 4 . Filter 
wash with water, discard the filtrate and pour ammonium 
acetate solution through the filter. K 2 Cr 0 4 will precipitate 
PbCr 0 4 from the filtrate. 

Lead sulphate yields ions thus: 

Pbso,<=tpb +2 +sor 2 . 



REACTIONS OF THE BASE-FORMING CONSTITUENTS 37 


The Pb +2 ions are probably taken up by the NH 4 Ac with the formation 
of complex acetate ions, whereby the above reaction continues to the 
right, and the lead sulphate dissolves. On the addition of K 2 Cr 0 4 
PbCr 0 4 is precipitated because lead chromate is far more insoluble in 
water than lead sulphate, the Cr 0 4 -2 ions take up the small number 
of Pb +2 ions yielded by the acetate complex, and the following equilib¬ 
rium tends to be displaced to the right: 

Pb 2 +Cr 0 r 2 <=±PbCr 0 4 . 

PbCr 0 4 is insoluble in acetic acid, but soluble in HC 1 
and in NaOH, sodium plumbite being formed in the latter 
case. 

8. KI precipitates Pbl2, soluble in large excess of the 
reagent: 

PbI 2 + 2KI^K 2 PbI 4 . 

SILVER 

17. Silver has three oxides, Ag 4 0 , Ag 2 0 , and Ag 2 0 2 , of 
which argentic oxide, Ag 2 0 , is of common occurrence. The 
base AgOH has not been isolated, but many of its salts of 
the general formula AgX are fairly stable. They have in 
aqueous solution the colorless, monovalent ion Ag +1 , and 
have the power under certain conditions of forming complex 
ions, as will be seen. Metal complexes of the type Ag 2 +1 are 
known, such as are found in the subchloride Ag 2 Cl. 

1. Heat a dry silver salt with Na 2 C03 on charcoal before 
the blowpipe. 

All silver compounds yield a metallic button under these conditions. 

2. KOH or NaOH precipitates Ag 2 0 from solutions of 
silver salts. The precipitate is soluble in NH 4 0 H (see next 
experiment). 

3. NH 4 OH precipitates AgOH, which changes largely into 


38 


QUALITATIVE CHEMICAL ANALYSIS 


Ag 2 0. The precipitate is soluble in excess NH 4 OH. HC1 
precipitates AgCl from the ammoniacal solution. 

The formation of Ag 2 0 may be expressed thus: 

AgN0 3 +NH 4 0H -> Ag0H+NH 4 N0 3 , 
2 Ag0H^Ag 2 0+H 2 0. 

The precipitate dissolves in excess NH 4 OH, due to the formation of a 
complex cation, Ag(NH 3 ) 2 +1 : 


AgOH<=±Ag +1 +OH-\ 


NH 4 0H<=>NH 3 +H 2 0, 

Ag +1 +2NH 3 <=> [Ag(NH 3 ) 2 ]+\ 

AgOH + 2NH 4 OH Ag(NH 3 ) 2 OH+ 2 H 2 0. 


o.r 


The complex silver ammonium ion is the more stable the more am¬ 
monia is present, as can be seen from the equilibrium equation 


Ag +1 +2NH 3 ^[Ag(NH 3 ) 2 ] +1 , 

Ag +1 X(NH 3 ) 2 

[Ag(NH 3 ) 2 ] +1 


or 


Now if something is added which removes NH 3 the above reaction will 
proceed to the left, and the complex ion will disappear. This is the 
condition when HC1 is added and AgCl is precipitated, since the NH 3 
is converted into NH 4 C1. 

The power of ammonia to form complex ions with silver (and other 
metals) may be due to simple addition. Just as water adds, with the 
formation of NH 4 OH, 



II 


H 

OH 


AgOH may add thus: 





REACTIONS OF THE BASE-FORMING CONSTITUENTS 39 


Such suppositions are based on the well-known property of nitrogen of 
passing from a valence of three to five. 

The formation of other complex ions may be considered on the 
same basis. It is known that chlorine readily passes from one valence 
to another, and the existence of complex anions in such salts as 
K 2 HgCl 4 may be explained on this power of chlorine to increase its 
valence. 

4. Na2CC>3 produces a precipitate of Ag2C03 which on 
heating decomposes into Ag 2 0. 

5. HC 1 precipitates AgCl, soluble in ammonia. 

From the solubility product 


Ag +1 X Cl -1 = K, 


there are a few Ag +1 ions in solution which with ammonia react to form 
the complex salt, silver-ammonium chloride: 

Ag +1 +Cl -1 + 2 NH 3 <=► Ag(NH 3 ) 2 +1 +C 1 - 1 . 

The explanation of the solubility of AgCl in excess ammonia and its 
precipitation from an ammoniacal solution by excess HC 1 should be 
thoroughly understood. 

AgCl dissolves in excess cone. HC 1 , alkali chlorides, and in KCN 
due to the formation of complex ions. Thus with KCN, KAg(CN) 2 
is formed. Formulate the equation which would express the relation 
between the concentration of the silver ions, cyanide ions, and complex 
ions at equilibrium, in this salt. 

6. H 2 S precipitates Ag 2 S, insoluble in NH 4 OH. 

Explain the fact that Ag 2 S is insoluble in ammonia, whereas AgCl 
is soluble. From the table of solubilities (p. 160) the solubility of 
Ag 2 S is 0.116 mg. or 0.0146 g. per liter; this corresponds to a concentra¬ 
tion of 0.0146/248 or 0.01624 mol per liter. Assuming complete dissoci¬ 
ation of the dissolved Ag 2 S (assume this in general in calculating solu¬ 
bility products unless otherwise directed) in the reaction 


Ag 2 S 2Ag +1 +S 2 , 


40 


QUALITATIVE CHEMICAL ANALYSIS 


from the solubility product. 

[Ag + Tx[S -*] = K, 

we have 

[Ag +1 ] = 2 X o.Oie 24 = 0.01648; [S“ ? ] =o.Oi 6 24. 

The product may then be calculated, compared with the solubility prod¬ 
uct for AgCl calculated from its solubility in a similar manner, and 
the behavior of the two substances toward NH 4 0 H will be clear. 

7. K2C1O4 precipitates Ag2Cr04. 

8. KI precipitates Agl, which is not very soluble in am¬ 
monia. 

Calculate the solubility products for AgCl and Agl and explain the 
difference in behavior of AgCl and Agl toward ammonia. 

18. The student should now analyze, according to a 
systematic procedure, one or more solutions containing salts 
of the silver group, and this plan should be followed in the 
study of the other groups. The metals are divided into vari¬ 
ous groups according to their power to be precipitated by 
a certain group reagent. Thus it has been observed by 
the student that the chlorides' of the metals so far studied 
are insoluble in water (as well as in dilute acids). The 
chlorides of other metals are soluble so that HC1, the group 
reagent for the silver group, precipitates only chlorides of 
mercurous mercury, lead, and silver. The solution obtained 
by filtering from these chlorides may contain salts of other 
metals, and in the regular procedure for the analysis for all 
the basic constituents should be preserved. The groups as 
they are precipitated must be separated into their constit¬ 
uents and the individual bases identified. 

The base-forming constituents are divided into five groups: 

Group I contains those bases whose chlorides are insoluble 
in dilute acids. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 41 

Group II contains those bases whose chlorides are soluble, 
but whose sulphides are insoluble in dilute hydrochloric 
acid. 

Group III contains those bases whose chlorides and sul¬ 
phides are soluble in dilute acids, but whose sulphides are 
either insoluble in ammonium hydroxide, or by hydrolysis 
are converted into hydroxides insoluble in ammonium hy¬ 
droxide. 

Group IV contains those bases whose chlorides and sul¬ 
phides are soluble in acids, and in alkalis, but whose car¬ 
bonates are insoluble in ammonium hydroxide in the presence 
of ammonium salts. (Magnesium does not fall in this class, 
but is included in the group for reasons to be stated.) 

Group V contains those bases whose chlorides, sulphides, 
and carbonates are all soluble in water. (CaS is insoluble in 
water, but by hydrolysis yields a soluble sulphydrate.) 

Tabulated outlines for the analysis of the groups are 
presented on pp. 147-153, and details of the procedure for 
the analysis of Group I are given in the following. The 
tables may be used to advantage after important details 
are mastered. The amount of precipitate formed in any 
test should be carefully observed, since the qualitative analyst 
must distinguish between large amounts and traces of con¬ 
stituents found. In the following analysis of unknown solu¬ 
tions, the amounts of the different constituents should be 
known by the instructor, and the student should endeavor 
to estimate the relative quantities present. 

Some tabular form of record of analyses acceptable to the 
instructor should be adopted. 


42 QUALITATIVE CHEMICAL ANALYSIS 

ANALYSIS OF GROUP I 

19 . Add to the unknown solution (which may be acid 
with HNO3) HC 1 until the precipitate ceases to increase in 
amount, and then add a slight excess of the reagent. 

In solutions which have a large amount of bismuth or antimony, 
BiOCl or SbOCl respectively, may be precipitated, but these oxy- 
salts are soluble in excess HC 1 . 

In the solutions given out as unknowns in this laboratory, not 
more than 500 mg. of a metal will be present in 20 c.c.; the metals 
may be present.in as low a concentration as 1 mg. in this volume. 
Approximately 20 c.c. of solution should, therefore, be taken for analysis. 
When unknown solids are analyzed, usually at least 1 mg. of metal 
will be present in the solution, the preparation of which is described 
in the latter part of this book. Detection of even smaller amounts, 
however, is not excluded.’ 

In adding the group precipitant, add HC 1 drop by drop. If there 
is a precipitate (presence of Group I) shake the solution after each 
addition, allow to settle, and note whether further addition brings 
down more precipitate. Finally add 0.5 c.c. HC 1 if the solution was 
originally acid, and 4 c.c. in excess if it was not. The excess acid 
dissolves basic salts and gives the proper concentration of acid for the 
precipitation of Group II. If Group I is absent, add no excess HC 1 
or 5 c.c. excess according to the reaction of the original solution. 

Could any other chloride than HC 1 be used to precipitate the metals 
of this group? What limitations would the use -of NaCl place on a 
systematic procedure for the detection of all the metals? 

Filter and wash the residue once with cold water. Save 
the combined filtrate and washings for the analysis for the 
remaining groups, if so desired. 

Lead chloride will remain in the filtrate to a relatively large extent 
on account of its solubility, but provision is made for the detection of 
lead in both residue and filtrate. 

20 . Pour boiling water several times through the filter con¬ 
taining the precipitated chlorides. Add 5 c.c. H2SO4 to the 
filtrate, which may contain lead chloride. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 43 

If barium salts had been carried through to this point they would 
be precipitated as BaS 0 4 and this would be mistaken for PbS 0 4 , so that 
a confirmatory test must be applied. 

Filter and wash the residue thoroughly with hot water 
to remove H2SO4, pour NH4C2H3O2 repeatedly over the 
residue on the filter; acidify the filtrate with acetic acid, 
and add K2Cr04. A yellow precipitate of PbCr04 is a test 

for lead. 

Why is PbS 0 4 soluble in ammonium acetate? Why must H 2 S 0 4 be 
removed before adding ammonium acetate? BaS 0 4 is more insoluble in 
water than PbS 0 4 and is not dissolved by the ammonium acetate on 
account of its low solubility product and probably since it has less 
tendency to form a complex ion. 

21 . Pour NH4OH through the filter containing the residue 
from the hot water treatment in 20 . A black residue of 
NH 2 HgCl plus Hg is a test for mercurous mercury. 

Do not confuse a brownish stain on the filter caused by the presence 
of certain other salts such as iron, with the black of the mercury. 

22. Acidify the NH4OH filtrate from 21 with HNO3 (test 
with litmus paper). A white precipitate of AgCl is a test 

for silver. 

Group II. A. The Copper Group. Hg +2 ; Pb +2 ; Cu +2 , 

Bi +3 , Cd +2 . 

Reactions 
MERCURIC SALTS 

23 . Mercuric salts may be formed from mercurous salts 
by oxidation: 

2Hg +1 *=* 2Hg+ 2 . 

The base Hg(OH) 2 is unstable and weak, so that many salts 


44 


QUALITATIVE CHEMICAL ANALYSIS 


of this series are strongly hydrolyzed. The halides are in 
general more soluble than those of the mercurous salts. 

Mercuric chloride seems to offer an exception to the theory 
of electrolytic dissociation, since it is very little ionized in 
aqueous solution. This fact is emphasized by its great 
stability. Thus it is scarcely affected by strong nitric acid, 
and from boiling cone. H2SO4 it can be volatilized unchanged. 

Mercuric chloride (also the bromide, iodide, and cyanide) 
tends to form salts of complex ions. Thus it is far more 
soluble in hydrochloric acid than in pure water, and salts 
of the general formula ^HCl-yHgCb result. The prac¬ 
tically neutral HgCl 2 is associated with the chlorine ions and 
the equilibrium may be illustrated by the following: 

3H+ 1 +3CH +HgCl 2 HsHgCls ^ 3 H +1 + [HgCl 5 ]- 3 . 

Mercuric chloride solutions react acid as the result of 
hydrolysis, but they may become neutral by the addition of 
alkali salts: 

2KCl+HgCl 2 K 2 HgCl 4 *± 2 K+ 1 +[HgCl 4 3 - 2 . 

This salt with a complex ion is not hydrolyzed to any degree 
because the ions are those of a strong base and acid. (In 
the previous example, the strength of hydrochloric acid is 
not affected by the addition of the neutral sublimate.) 

1. Heat corrosive sublimate, HgCl 2 , in a closed tube. 

2. KOH precipitates HgO from solutions of mercuric salts. 

3. NH 4 OH precipitates NH 2 HgCl from solutions of HgCl 2 , 
and NH 2 HgN03-Hg0 from solutions of Hg(NOs) 2 . 

4. Na 2 C03 produces in solutions of mercuric salts a pre¬ 
cipitate of a basic carbonate, which yields HgO when the 
solution is boiled. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 45 


5. HgS is precipitated by H2S, and is insoluble in HC 1 , 
and in dilute HNO3. It is relatively very insoluble in water 
and therefore yields but few ions when suspended in it, a 
fact that will explain later the differential effect of acid upon 
copper and other sulphides. Aqua regia dissolves HgS very 
readily, due to the action of free chlorine produced by the 
reagent. 

Aqua regia is prepared by mixing three volumes of cone. HC 1 with 
one volume of cone. HN 0 3 . The nitric acid oxidizes the hydrochloric 
acid and chlorine is thereby evolved. 

When cone. HN 0 3 oxidizes cone. HC 1 the following assumptions may 
be made if the reaction is to be written in the ionic form. The graphic 

//° 

formula for nitric acid may be expressed by H—O—N/ or, if hydrated 

X) 

as in the normal acid, N(OH) 5 . Now as with Pb(OH) 2 ( 16 , 4) the 
hydroxide N(OH) 5 may be assumed to be amphoteric, although there 
is no direct experimental evidence to corroborate this assumption. In 
the presence of the strong acid HC 1 , the acid dissociation 

HN 0 3 • 2 H 2 0 H +1 -f N 0 3 -1 + 2 H 2 0 

would be suppressed by the common-ion effect. The dissociation as a 
base would be more probable: 

N(OH) a N +5 + 5 OH _1 , 

and the reaction between cone. HC 1 and cone. HN 0 3 would be expressed 
as follows: 

N(OH) 5 +2HCI -> N(OH) 3 + 2 H 2 0 +Cl 2 , 

or 

N +5 +5(OH) _l + 2 H +1 +2CI -1 —> N +3 + 3(OH) -1 + 2H 2 0+C 1 2 . 

But N(OH) 3 is hydrated nitrous acid, H—O—N== 0 , the acid chloride 
of which, NOC 1 or nitrosyl chloride (an acid chloride is a derivative of 
an acid in which chlorine has been substituted for an hydroxyl group) 
is readily formed: 

N +3 ^(OH)- 1 +H +1 +C 1- 1 -> NOC 1 + 2 H 2 0 . 


46 


QUALITATIVE CHEMICAL ANALYSIS 


The net result of the reaction 

HN 0 3 • 2H 2 0 +3HCI -> N 0 C 1 + 4 H 2 0 +C 1 i, 

is the lowering of the valence of chlorine from minus one to zero. 
Or one nitrogen has taken up two electrons from two chlorines. 

The following equations may explain the oxidation according to the 
dualistic basis: 

2 HN 0 3 -+H 2 0 +N 2 0 5 , 

N 2 0 5 -> 2 NO+ 30 , 

6HCI+3O -> 3 h 2 o+xi 2 , 

Cl 2 +2NO —2NOCI, 

2 HNO 3 + 6 HCI -> 2N0C1+2C1 2 + 4 H 2 0, 
or 

HNO 3 + 3 HCI -^N0C1+C1 2 +2H 2 0. 

6. KI precipitates yellow Hgl2 from sublimate solutions. 

This yellow form, however, is in an unstable condition (it is stable 
above 126° C.) and changes very quickly into the red form. Hgl, dis¬ 
solves in excess KI, forming the complex salt K 2 HgI 4 . This salt is stable 
in the presence of KOH, and such an alkaline solution (Nessler’s reagent) 
is used as a test for ammonia (which see). 

7. SnCl2 reduces mercuric salts first to calomel, and then 
to metallic mercury. 

If the solution of the mercury salt is cold, and cold stannous chloride 
is slowly added, the formation of the mercurous compound may be clearly 
observed. 


BISMUTH 

24 . Bismuth has a valence of three and five, and has two 
oxides, yellow Bi 2 C>3 and brown Bi 2 0 5 . Bi 2 0 5 is the anhy¬ 
dride of an acid HBiOs. Bi 2 03 is the anhydride of the base 
Bi(OH) 3 , which yields salts with the colorless ion Bi +3 . 
These salts are largely hydrolyzed in water and are kept in 
solution only by addition of the corresponding acid. 



REACTIONS OF THE BASE-FORMING CONSTITUENTS 47 


1. Heat a bismuth compound with Na2C03 on charcoal 
before the blowpipe and note the character of the metallic 
button and of the incrustation. 

2. NH4OH precipitates Bi(OH) 3 . Filter, dissolve the 
residue in a small amount of cone. HC 1 and pour the solu¬ 
tion into a large amount of water. BiOCl is precipitated. 
Add an excess of HC 1 , and explain the result. 

3. KOH precipitates Bi(OH) 3 . Add bromine water and heat. 

The precipitate changes slowly to a brown color due to the forma¬ 
tion of bismuthic acid, HBi0 3 . 

Bi(0H) 3 +2K0H+Br 2 -^HBi0 3 +2KBr + 2H 2 0, 
or 

Bi +3 + 2 Br —> Bi +5 + 2 Br _1 . 

4. Na2C0 3 precipitates basic salts, e.g., Bi 2 0 3 -C02. 

5. H2S produces a precipitate of Bi 2 S 3 , insoluble in dilute 
HC 1 , soluble in hot cone. HC 1 and in hot dilute HN 0 3 . 
(Cf. copper.) 

6. Na 2 Sn02 in excess reduces bismuth salts (and Bi(OH) 3 ) 
to metallic bismuth. 

Na 2 Sn0 2 (cf. tin) is prepared by adding NaOH to SnCl 2 until the 
precipitate first formed goes into solution, which should be clear. The 
sodium stannite is oxidized by the bismuth compound to sodium stannate, 
Na 2 Sn0 3 . 


COPPER 

25. Copper may have a valence of one or two in its 
compounds, and has two oxides, red cuprous oxide, Cu 2 0 , 
and black cupric oxide, CuO, to which correspond the hydrox¬ 
ides and two series of salts. Solutions containing the cupric 
ion Cu +2 vary in color from green to blue (cf. Exp. 1 below) 
while the cuprous ion Cu +1 is colorless. (Cuprous salts are 


48 QUALITATIVE CHEMICAL ANALYSIS 

insoluble in water, but yield enough ions to verify this state¬ 
ment.) 

Cuprous salts are very unstable and oxidize in the air 
to cupric salts (with the exception of certain complex salts). 
They have the general formula CU2X2. Cuprous salts dis¬ 
solve in HC 1 and other chlorides with the probable formation 
of complex salts. The composition of these is variable; in solu¬ 
tion with KC 1 such salts as KCuCh are indicated. On account 
of the instability of cuprous salts and their rapid conversion 
into cupric salts, only the latter will be treated experimentally. 

1. Copper chloride serves as an example visually to observe 
ionization phenomena, since the undissociated molecule is 
green in color, whereas cupric ions are blue. 

Dissolve about 5 grams CuC 1 2 -2H 2 0 in 5 c.c. water and 
cautiously dilute until the solution has changed from green 
to blue. Add cone. HC 1 , drop by drop, and note that the 
green color is restored. 

Explain, and formulate the equation which expresses the equilibrium 
between the concentration of undissociated salt and its ions in a solu¬ 
tion of copper chloride. 

Prepare a solution of copper chloride, which is green in 
color, but which a few drops of water would change to blue. 
Add a few crystals of Hg(N 0 3 ) 2 , shake, and observe the 
change in color towards blue. 

Mercuric nitrate will react with copper chloride and form mercuric 
chloride, HgCl 2 , which is only slightly dissociated in solution into its 
ions. If chlorine ions are thereby removed from the solution, how 
may the above change in color be explained? 

Why was CuCl 2 • 2H.O, and not CuS 0 4 • 5H 2 0, selected for the above 
experiments? 

2. Heat a copper compound with Na 2 C 0 3 on charcoal 
before the reducing flame of the blowpipe. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 49 


All copper compounds yield red metallic copper, this color usually being 
hidden by a superficial layer of black oxide which should be scraped away. 

3. Make a small loop on the end of a platinum wire, heat 
and dip in microcosmic salt (sodium ammonium hydrogen 
phosphate) and heat until a transparent bead is obtained. 
Add a very small amount of a copper compound and heat in 
the oxidizing flame (tip of the non-luminous flame) and then 
in the reducing flame (tip of the inner blue cone) of the 
Bunsen burner. Examine the bead when cold and note the 
blue and red colors respectively. 

Microcosmic salt (NaNH 4 HP0 4 *4H 2 0) when heated alone loses 
water and yields transparent sodium metaphosphate NaP0 3 . CuS0 4 , 
heated in this bead, loses sulphur trioxide: 

CuS0 4 —> CuO-fS0 3 , 

and the copper oxide combines with the metaphosphate to form the 
blue sodium copper phosphate: 

CuO+NaP0 3 —>NaCuP0 4 . 

In the reducing flame this salt is broken up, and the following reaction 
may take place: 

2 NaCuP 0 4 +C -> 2NaP0 3 +Cu 2 0+C0, 

the red color being imparted to the reformed sodium metaphosphate 
by Cu 2 0. 

4. NH4OH precipitates from solutions of cupric salts a 
basic salt passing into the true base which readily dissolves in 
excess of the reagent: 

2C11SO4+2NH4OH -» CuS 0 4 - Cu(0H)2+(NH4) 2 S04, 

CuS 04 -Cu( 0 H) 2 + 2 NH 40 H-» 2 Cu( 0 H) 2 +(NH 4 ) 2 S 0 4 . 

When the ammoniacal solution of Cu(OH) 2 in excess ammonia is 
electrolyzed the copper migrates to the cathode, and also copper does 
not show a pronounced amphoteric character with great tendency 
to dissolve in strong alkali as does lead hydroxide. The existence of 


50 QUALITATIVE CHEMICAL ANALYSIS 

complex cations is thus suggested. In excess ammonia a stable copper- 
ammonium compound is formed thus: 

Cu(OH) 2 + 4 NH 4 OH Cu(NH 3 ) 4 (OH) 2 +2H 2 0. 

Since it is known that these complex metal ammonium hydroxides are 
more strongly basic than ammonium hydroxide, we may write: 

Cu(NH 3 ) 4 (0H) 2 +(NH 4 ) 2 S0 4 - Cu(NH 3 ) 4 S0 4 +2NH 4 0H. 

The cations of this copper-ammonium salt have a tendency to dissociate 
thus: 

Cu(NH 3 ) 4 +2 *=± Cu +2 + 4 NH 3 . • 

[Cu +2 ] x[NH 3 ] 4 
[Cu(NH 3 ) 4 +2 ] 

and we can thus, as with the corresponding silver complex, under¬ 
stand the measure of the stability of this cation. Addition of NH, 
will increase the numerator, whereby the complex ion must increase in 
amount and the Cu +2 must decrease to satisfy the above equilibrium 
ratio. Addition of acid removes NH 3 , the complex dissociates more and 
more, and the color of the solution which is the deep blue of the copper 
complex, changes to the lighter blue of the Cu +2 ions. 

H2S precipitates CuS from ammoniacal solutions of cupric 
salts. (See below.) 

KCN decolorizes ammoniacal cupric salt solutions due to the 
formation of potassium cuprocyanide. 

Copper is now in the anion, because copper migrates to the anode 
if the solution is electrolyzed. 

Cu(NH 3 ) 4 + 2 -► Cu+ 2 + 4 NH 3 , 

NH 3 +H 2 0 -> NH 4 OH-> NH 4 +i +OH- 1 , 

KCN —»K +1 +CN~V 

2Cu+ 2 +CN- 1 +20H“ 1 -> 2Cu+ 1 +CN0- 1 +H 2 0, 

2 Cu+ 1 +2CN- 1 -* Cu 2 (CN) 2 , 
or 

(a) 2 Cu(NH 3 ) 4 S0 4 + 7 H 2 0+ 3 KCN Cu 2 (CN) 2 

-fKCN 0 +(NH 4 ) 2 S 0 4 +K 2 S 0 4 + 6 NH 4 0 H. 



REACTIONS OF THE BASE-FORMING CONSTITUENTS 51 
The cuprous salt forms a complex thus: 

2 Cu+ 1 +8CN- 1 -> 2 Cu(CN) 4 - 3 

or 

(b) Cu 2 (CN) 2 +6KCN —> 2 K 3 Cu(CN) 4 . 

The entire action of KCN on the copper ammonium salt may be 
expressed by adding (a) and ( b ). 

If H 2 S is passed into this solution containing the cuprous 
cyanide ion, no precipitate of copper sulphide is obtained, an 
evidence of the greater stability of this ion than of the copper- 
ammonium ion. 

5. KOH precipitates Cu(OH)2, changing on heating into 
CuO. 

Strongly concentrated alkali dissolves Cu(OH) 2 . Explain this action, 
bearing in mind the behavior of excess alkali on lead hydroxide. 

6 . Na 2 C03 precipitates a basic carbonate, e.g., Cu(OH) 2 * 

Q1CO3. 

7. H 2 S produces a precipitate of CuS which is insoluble 
in dilute HC 1 , but soluble in warm dilute HNO3. 

With CuS suspended in water: 

CuS <=* Cu+ 2 +S- 2 , 

and from the solubility product principle, 

Cu+ 2 XS~ 2 = A:. 

CuS is therefore precipitated from solutions of cupric salts when enough 
S -2 ions are present for the characteristic constant to be exceeded. H 2 S 
dissociates primarily thus: 

H 2 S <=^ H+ 1 +SH -1 , 

and in a saturated solution of H 2 S the amount of undissociated H 2 S 
will be constant and 


[H* 1 ] X [SH _1 ] = Ki. 


52 


QUALITATIVE CHEMICAL ANALYSIS 


The sulphydrate ion dissociates to a less extent: 

SH _1 H +1 +S -2 , 
or 

[H +1 ] X[S~ 2 ] y 
[SH- 1 ] Ki ' 

Multiplying the last two equations we have 
[H +1 ] 2 X[S- 2 ] = iV, 

which follows at once from the equation 

H 2 S^2H +1 +S“ 2 . 

When HC 1 is added the hydrogen ion content increases, and thereby 
the sulphide ion content is decreased as can be readily seen from 
the last two equations. Since CuS is insoluble in dilute HC 1 , the 
sulphide ions are not reduced below the number sufficient for the 
solubility product of CuS in water. But cone. HC 1 would decrease 
the sulphide ions very markedly; in fact CuS is soluble in this reagent, 
the sulphide ions being reduced below the amount necessary to satisfy 
the solubility product. The fact that CuS dissolves in dilute HN 0 3 
and not in dilute HC 1 must involve a further principle since HN 0 3 
is no stronger than HC1 and there can be no great difference in their 
action due to different amounts of hydrogen ion. When HN 0 3 dis¬ 
solves CuS, free sulphur is produced, due to oxidation by the HN 0 3 of 
the H 2 S formed bv the first action: 

CuS + 2 HN 0 3 ->Cu(N 0 3 ) 2 +H 2 S, 

3 H 2 S+ 2 HN 0 3 -mH 2 0 + 2 N 0 + 3 S. 

The sulphur ions are converted into free sulphur and thus removed 
from the field of influence and the CuS comes to complete solution. 

It was seen in Exp. 5 that H 2 S precipitates CuS from ammoniacal 
solutions. H 2 S combines with ammonia to form ammonium sulphide: 

2NH 3 +H 2 S -> (NH 4 ) 2 S, 

and as (NH 4 ) 2 S is more strongly dissociated than H 2 S, the sulphide 
ions are moie numerous: 

(NH 4 ) 2 S^±2NH 4 +1 +S- 2 . 



REACTIONS OF THE BASE-FORMING CONSTITUENTS 53 

The copper ions coming from the slightly dissociated complex copper- 
ammonium ions and the sulphide ions are sufficient for the solubility 
product of CuS to be exceeded and CuS is precipitated. (CuS is not 
completely insoluble in (NH 4 ) 2 S). 

The fact that Cu 2 S is not precipitated from KCN solutions of copper 
salts may be explained along the same lines as the above. The com¬ 
plex cuprous-cyanide ions do not yield enough copper ions to exceed 
the solubility product of Cu 2 S: 

[Cu+ 1 ] 2 x[S~ 2 ] = iU. 

Explain the fact that CuS is soluble in KCN. 

Calculate from solubility tables the solubility products for CuS 
and HgS and explain the difference in solubility in HN0 3 . 

8. K4Fe(CN)o precipitates Cu 2 Fe(CN) 6 , insoluble in acetic 
acid. 

CADMIUM 

26. Cadmium has a valence of two, with one oxide, 
brown CdO. The hydroxide Cd(OH) 2 yields salts with the 
colorless ion Cd +2 . 

1. Heat a cadmium compound with Na 2 CC>3 on charcoal 
before the blowpipe. All cadmium compounds yield a brown 
incrustation. 

2. NH4OH precipitates Cd(OH) 2 which dissolves in excess 
of the reagent, forming a complex cadmium-ammonium salt, 
e.g., Cd(NH 3 ) 4 S 0 4 . 

KCN converts such salts into salts of complex cadmium-cyanide 
anions—K 2 Cd(CN) 4 . From such solutions CdS is precipitated by H 2 S. 
Explain this action, comparing the stability of K 2 Cd(CN ) 4 with that 
of K 3 Cu(CN) 4 . 

3. KOH precipitates Cd(OH) 2 . 

4. Na 2 C03 precipitates CdC 0 3 . 

5. H 2 S produces a precipitate of CdS, insoluble in dilute 
HC 1 , soluble in hot cone. HC 1 and hot dilute HN 0 3 . 


54 


QUALITATIVE CHEMICAL ANALYSIS 


Group II. B. The Arsenic Group. As+ 3 , As+ 5 , Sb+ 3 , 
Sb +5 , Sn+ 2 , Sn +4 . 

Reactions 

ARSENIC 

27 . Arsenic has a valence of three and five, and two 
oxides, white arsenious oxide, AS2O3, and arsenic pentoxide, 
AS2O5. The hydroxides corresponding to AS2O3 have not been 
isolated, but salts of orthoarsenious acid, H3ASO3, meta- 
arsenious acid, HAs 0 2 , and pyroarsenious acid, H4AS2O5, are 
known. AS2O3 is not very soluble in water, but dissolves 
readily in alkali hydroxides or carbonates: 

AS2O3+3H2O 2As(OH)3, 

As(OH) 3 +3NaOH Na 3 As0 3 +3H 2 0. 

AS2O3 is pre-eminently an acid anhydride, but its ampho¬ 
teric nature may be illustrated by its reaction with hydro¬ 
chloric acid: 

As(OH) 3 + 3 HC 1 <=> ASCI3+3H2O. 

Arsenic chloride will be present in such solutions to an 
appreciable amount only with a large excess of HC 1 . 

Explain the action of HC1 on the basis of the suppression of the acid 
dissociation of As(OH) 3 . 

AS2O5 is the anhydride of the arsenic acids: ortho, H3ASO4, 
meta, HASO3, and pyro, H4AS2O7. AS2O5 dissolves readily in 
water, forming orthoarsenic acid: 

AS2O5+3H2O ^ (HO) 3 AsO. 

Since arsenic acid is stronger than arsenious acid, its ampho¬ 
teric nature is less in evidence, and AsC 1 5 has not been 




REACTIONS OF THE BASE-FORMING CONSTITUENTS 55 


isolated, although its existence is very probable. The alkali 
salts are soluble. 

Arsenic ions As+ 3 and As+ 5 , and the acid ions such as 
As 0 3 " 3 and AsC>4 -3 are colorless. 

1. Heat an arsenic compound with Na 2 CC>3 on charcoal. 

All arsenic compounds give off a garlic odor and very volatile arsenious 
oxide is formed. 

2. H 2 S precipitates As 2 S 3 from a hydrochloric acid solu¬ 
tion of arsenious acid. 

This may be explained on the basis of the basic dissociation of 
arsenious acid, the acid dissociation being suppressed by HC 1 . Note, 
however, that the HC 1 suppresses the dissociation of H 2 S, but not 
enough to prevent the precipitation of very insoluble As 2 S 3 : 

2As +3 +3S -2 -> As 2 S 3 . 

As 2 S 3 , the sulphur substitution product of As 2 0 3 , is soluble in ammonium 
carbonate, alkali hydroxides and sulphides. With colorless ammonium 
sulphide (NH 4 ) 2 S, it forms the soluble salt of a sulpho acid, H 3 AsS 3 , 
analogous to arsenious acid H 3 As 0 3 . 

3(NH 4 ) 2 S+As 2 S 3 2(NH 4 ) 3 AsS 3 . 

If ammonium polysulphide, (NH 4 ) 2 Sx, however, acts on As 2 S 3 , instead 
of ammonium sulpharsenite, ammonium sulpharsenate is obtained, due 
to the oxidation of As 2 S 3 by the excess sulphur in the polysulphide: 

3(NH 4 ) 2 Sx+As 2 S 3 -> 2 (NH 4 ) 3 AsS 4 +(32—3) S. 

HC 1 yields a precipitate of As 2 S 3 with the ammonium sulpharsenite 
solution and As 2 S 5 with the sulpharsenate solution on account of the 
instability of the sulphoacids. Thus: 

2 (NH 4 ) 3 AsS 3 + 6 HC 1 -> 2H,AsS 3 +6NH 4 Cl, 

2H 3 AsS 3 —»As 2 S 3 +3H 2 S. 

3. H 2 S converts arsenious acid (in the absence of other 
electrolytes) into colloidal As 2 S 3 . Pour the colored mixture 




56 QUALITATIVE CHEMICAL ANALYSIS 

through a filter. Add a little HC 1 and note the precipitation 
of AS2S3. 

Although the colloidal As 2 S 3 appeared to be in true solution, by the 
use of the ultra-microscope distinct solid particles would be indicated. 
Colloidal mixtures, presenting the appearance of ordinary solutions, 
can nevertheless usually be shown to be heterogeneous that is, there 
are two distinct phases present. Two classes are known, emulsoids 
and suspensoids. Emulsoids are those colloidal mixtures in which the 
two phases are both liquids; gelatine solutions are a type of this class. 
Suspensoids consist of solid particles in a liquid medium, the particles 
being larger aggregates than those present in the usual molecular solu¬ 
tions. Colloidal mixtures cannot be shown to be heterogeneous by the 
ordinary methods of filtration, but may be separated by the use of 
animal membranes, by parchment paper, collodion membranes, etc. 
Colloids such as As 2 S 3 bear a negative electric charge, the surrounding 
medium being positively charged. This charge seems to be closely 
associated with the colloidal condition and the removal of it destroys 
that condition. Thus by adding HC 1 , or other electrolytes such as 
salt solutions, the solid particles combine to form larger aggregates 
and As 2 S 3 is precipitated. This precipitation is the result of the neu¬ 
tralization of the negative charge of the colloid by the positive charge 
of the cation of the electrolyte, and it is found that the precipitating 
power of the cations of electrolytes is a function of the valence of the 
cations. Bivalent cations have greater precipitating power than uni¬ 
valent cations. Positive colloids are known, e.g., Fe(OH) 3 , prepared 
by forming the hydroxide in the absence of electrolytes. The precipitation 
of such colloids will depend on the valence of the anions of added salts. 

All substances probably have the power of existing in the colloidal 
condition so that for the purposes of precipitation and filtration such 
a condition must be avoided, and it is always desirable to have elec¬ 
trolytes present. Heating favors precipitation as well, so that such 
colloidal mixtures as silver chloride in water, sulphides, etc., are pre¬ 
cipitated in hot solution in the presence of elctrolytes. Substances 
with a great tendency to go into the colloidal condition should not be 
washed by pure water, but by aqueous solutions of salts, acids, or bases, 
according to the nature of the colloid. 

4. AgNC>3 precipitates Ag 3 As03 from solutions of potassium 
arsenite. 




REACTIONS OF THE BASE-FORMING CONSTITUENTS 57 


5. H2S precipitates AS2S3 from cold hydrochloric acid solu¬ 
tions of potassium arsenate only very slowly. 

H3ASO4+H2S —> H3ASO3 -j-S -I-H2O, 
2H3ASO3+3H2S —> AS2S3+6H2O. 

With considerable excess HC1, H 2 S precipitates As 2 S besides As 2 S 3 , 
and the precipitation proceeds faster than in the cold. The formation 
of As 2 S 5 may be explained by the weakly amphoteric nature of arsenic 
acid: 

H 3 As0 4 +H 2 0 <=± As+«+ 5 OH- 1 , 
and the formation of arsenic pentachloride: 

As(OH) 5 + 5 HCI <=> AsCh+ 5 H 0 O. 

As 2 S 5 dissolves in alkah sulphides to form sulpharsenates: 

3(NH4)2S+As 2 S, 2(NH 4 ) 3 AsS4, 

from which solutions As 2 S 5 is reprecipitated by HC1: 

2 (NH 4 ) 3 AsS 4 + 6 HCl -> 2 H 3 AsS 4 + 6 NH 4 CI, 

2 H 3 AsS 4 —> As 2 S 5 + 3 H 2 S. 

6. AgN03 precipitates Ag3As04 from solutions of potassium 
arsenate. 

7. Magnesia mixture (MgC^+NEUCl) precipitates mag¬ 
nesium-ammonium arsenate from ammoniacal solutions of 
arsenates. 

Explain the action of the NH 4 C1. 

ANTIMONY 

28. Antimony has a valence of three and five, with three 
oxides, white antimony trioxide, Sb203, and antimony tetroxide, 
Sb 2 0 4 , and yellow antimony pentoxide, Sb20s. These form 
amphoteric hydroxides, although Sb203 is markedly basic 


58 QUALITATIVE CHEMICAL ANALYSIS 

in its tendencies. (Cf. periodic system.) Sb2C>3 yields prob¬ 
ably three acids, ortho, H 3 Sb 0 3 , meta, HSb 0 2 , and pyro- 
antimonous acid, H4Sb20s. Sb20s also yields three acids, 
ortho, H 3 Sb04, meta, HSb 0 3 , and pyroantimonic acid, H 4 Sb207. 
The chlorides SbCl 3 and SbCl 5 are known and hydrolyze 
largely in water. The ions Sb +3 , Sb +5 , SbC>2 _1 , Sb 0 4 -3 , etc., 
are colorless. 

1. Heat an antimony compound with Na2C0 3 on charcoal. 
Note the metallic button and the incrustation. 

2. Water precipitates antimony oxychloride from anti¬ 
mony trichloride: 

SbCl 3 +H 2 0 <=> SbOCl + 2HCl. 

Filter and add to the residue a solution of Rochelle salt 
(potassium sodium tartrate). The SbOCl dissolves, forming 
tartar emetic (potassium antimonyl tartrate): 

KNaC 4 H 4 0 6 +Sb 0 Cl K(Sb0)C4H 4 0 6 +NaCl. 

3. H 2 S precipitates Sb2S 3 from solutions of SbCl 3 . 

Sb 2 S 3 is soluble in (NH 4 ) 2 S, forming ammonium sulphantimonite, 
(NH 4 ) 3 SbS 3 , and in (NH 4 ) 2 S;r, forming ammonium sulphantimonate, 
(NH 4 ) 3 SbS 4 . From these solutions Sb 2 S 3 and Sb 2 S 5 , respectively, are 
precipitated by dilute HC1. Explain these reactions, which are similar 
to those of arsenic. The sulphides are soluble in hot cone. HC1, yield¬ 
ing SbCl 3 in both cases. 

4. Iron (pure wire may be used) precipitates metallic 
antimony from HC 1 solutions of SbCl 3 . Filter and dissolve 
the precipitate in tartaric acid and a little nitric acid. H2S 
will precipitate Sb 2 S 3 from this solution. 

5. Immerse in an HC 1 solution of SbCl 3 , platinum foil on 
which is placed a piece of tin. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 59 


Tin displaces the dissolved antimony by galvanic 
action. Note the relative positions of iron, tin, hydro¬ 
gen, and antimony in the accompanying table pre¬ 
senting some of the common elements arranged 

according to the electrochemical series. 

The degree to which the ions of an element tend 
to retain their charge is called the electro-affinity of 
the element. The ions of potassium, sodium, fluor¬ 
ine, and chlorine have little tendency to give up their 
charges, while the ions of mercury, gold, boron, and 

silicon readily lose their charge. The former are 

called strong ions and the latter weak ions. Zinc 

has a stronger electro-affinity than copper, and zinc 
precipitates copper from solutions of its salts—the 
charge of the copper ions are transferred to the zinc 
jons: 

Zn+Cu +2 —> Zn +2 +Cu. 

Chlorine has a stronger electro-affinity than iodine 
and will displace it from solutions of its salts: 

Cl 2 +2l- 1 -^2Cr i +l2. 

I 

The student should consult other texts for a more 
complete discussion of this subject. 


a 

s 

Ih 

<2 

o 

§ 

m 


Potassium 

Sodium 

Lithium 

Barium 

Strontium 

Calcium 

Magnesium 

Aluminium 

Chromium 

Manganese 

Zinc 

Cadmium 

Iron 

Cobalt 

Nickel 

Tin 

Lead 

Hydrogen 

Antimony 

Bismuth 

Arsenic 

Copper 

Mercury 

Silver 

Platinum 

Gold 


oj 

W 

9 

| 

c 

'd 

o 

< 


Silicon 

Carbon 

Boron 

Nitrogen 

Phosphorus 

Sulphur 

Iodine 

Bromine 

Chlorine 

Oxygen 

Fluorine 


6. Add HC 1 and KI to a solution of potassium pyroanti- 
monate, K2H2Sb2C>7. 


HI reduces antimonic compounds (not antimonous compounds) with 
the separation of free iodine. In the above test, SbCl 5 is first formed 
and then reduced: 


Sb +5 + 2l- 1 ->Sb +3 +I,. 

7. Na2CC>3 precipitates a more or less hydrated oxide of 
antimony from solutions of SbCl3. The action of alkali 
hydroxides is entirely analogous. * 





60 


QUALITATIVE CHEMICAL ANALYSIS 


TIN 

29 . Tin has a valence of two and four and has two oxides, 
black stannous oxide, SnO, and white stannic oxide, SnC>2. 
The hydroxides are known and are amphoteric, being more 
asic, however, than either the arsenic or antimony com- 

unds. The hydrates of SnC>2 exist in two forms, normal 
stannic acid, Il2Sn03, and metastannic acid, HioSnsChs. 
'""he chlorides, SnCh and SnCU, are quite stable in aqueous 
solution. The ions Sn +2 , Sn^, Sn02“ 2 , and Sn03 -2 , are 
colorless. 

Heat a tin compound with Na2C03 (if no reduction is 
observed, add KCN) on charcoal. Note the metallic button 
and the incrustation. 

2. Add KOH to a solution of SnC^ until a precipitate of 
Sn(OH)2 is formed. Test the solubility of the stannous 
hydroxide in KC 1 and in excess KOH. Explain these reac¬ 
tions. SnCb and K2Sn02 are formed respectively. Repeat 
the experiment with SnCU; normal stannic acid, H2Sn03, 
being precipitated by a limited amount of KOH. 

3. Na^CCs precipitates Sn(OK) 2 and H 2 Sn03 from solu¬ 
tions of SnCb and SnCU respectively. The precipitates do 
not dissolve readily in excess of the reagent. 

4. K 2 S precipitates SnS from SnCl2, and SnS2 from SnCU. 
SnS is insoluble in (NH^S (difference from arsenic and anti¬ 
mony) but is soluble in (NH^S*, forming ammonium sul- 
phostannate, (NH^SnSg. From this solution SnS2 is pre¬ 
cipitated by HC 1 . SnS2 is soluble in both the colorless and 
the yellow ammonium sulphides. 

Explain these differences in the solubilities of the two sulphides on 
the beysis of the greater basic tendency of stannous compounds, 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 61 

5. Reduce some SnCU to S11CI2 with iron and HC1 and 
add the clear solution to HgCl2. 

6. Heat metallic tin with cone. HNO3. Metastannic acid 
is formed. This is acted on by hot cone. HC 1 , and water 
converts the resulting product into a soluble acid chloride, 
Sn 5 0 5 Cl2(0H) 8 . 


ANALYSIS OF GROUP II 

30 . Precipitation of Group II. Saturate the cold filtrate 
from Group I ( 19 ) with H2S. Filter and wash with hot 
water until the residue is free from acid (test washings with 
litmus paper). 

If the solution is insufficiently acid, metals of Group III, especially 
zinc, may be precipitated as sulphides. They are not precipitated in 
solutions of the recommended concentration of acid, since their sulphides 
are more soluble in water than those of the copper and arsenic groups; 
there are insufficient sulphide ions, which have been decreased by the 
H +1 ions of the added acid, for the solubility product of ZnS, e.g., 
([Zn +2 ] X[S~ 2 ] = K) to be exceeded. 

A white precipitate of sulphur may be obtained by passing H 2 S 
into solutions containing oxidizing agents such as cone. HN 0 3 , KMn 0 4 , 
K 2 Cr 2 0 7 , ferric salts, etc. 

The colors of the sulphides of Group III are as follows: black or 
brownish-black, HgS, PbS, Bi 2 S 3 , CuS, SnS; yellow, CdS, As 2 S 3 , As 2 S 6 , 
SnS 2 ; orange, Sb 2 S 3 ,. Sb 2 S 5 . 

Pass H 2 S into the filtrate maintained at a temperature 
of about 90° for ten minutes. If there is a precipitate, con¬ 
tinue passing in H2S for about twenty minutes longer. Filter 
and wash. 

Arsenic in the pentavalent condition is best precipitated under these 
conditions—that is, in a fairly strong acid solution kept at a temperature 
a little below that of boiling. 

Dilute the filtrate (or the solution in which H2S has failed 
to bring down a precipitate) to 100 c.c. (see note below) and 


62 


QUALITATIVE CHEMICAL ANALYSIS 


pass H 2 S into the cold solution to saturation. Filter and 
wash any precipitate. 

After diluting to ioo c.c., and before passing in H 2 S the last time, 
it is advantageous for the student to make the following test sug¬ 
gested by Stieglitz. 1 A drop of the solution is placed on methyl violet 
paper, and if a blue or blue-green color is obtained, the solution is 
dilute enough to precipitate all the copper group sulphides. If, however, 
a yellow-green to yellow tint is obtained, the concentration of the acid 
is too great for the complete precipitation of the more soluble sulphides 
(cadmium, lead). Dilute the solution until the test shows the blue- 
green to blue color, pass in H 2 S, and if a precipitate forms, filter and 
wash. Methyl violet is an indicator for the following strengths of acid: 
o.i-N HC 1 , blue; 0.25-N HC 1 , blue-green; 0.33-N HC 1 , yellow-green; 
0.5-N HC 1 , yellow. A concentration corresponding to 0.25-N HC 1 is 
the proper concentration of hydrogen ion to ensure the complete pre¬ 
cipitation of all the sulphides; it must be at least 0.1-N to prevent 
the precipitation of Group III sulphides. HgCl 2 interferes with the 
methyl violet test, but this is made after the very insoluble HgS has 
been removed by the first precipitation with H 2 S. 

31 . Separation of the Copper and Arsenic Groups. A 

small part of the combined H 2 S precipitates obtained as in 
30 is warmed with a little dilute (NH^S*. (a) If complete 

solution takes place, the copper group is absent and the entire 
H 2 S precipitate may be tested for the arsenic group as in 37 . 
( b ) If there is a residue after treating with the polysulphide, 
filter, and make the filtrate faintly acid with HC 1 . If the 
precipitate is pure white (sulphur) the arsenic group is absent 
and the entire H 2 S precipitate may be tested for the copper 
group as in 32 . (c) If there is a residue after treating the 

H 2 S precipitate with the polysulphide, and the acidified filtrate 
is yellow to orange, both the copper and arsenic groups are 
present. Warm the entire H 2 S precipitate with 10 c.c. full 
strength (NH^S* for five minutes, dilute with an equal 
1 Elements of Qualitative Analysis. II, pp. 102 and 103. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 63 

volume of water, filter, and wash with hot water. Treat 
the residue for the copper group according to 32. Make the 
filtrate faintly acid with HC1, filter by suction, and treat 
the ccmparatively dry residue for the arsenic group according 

to 37. 

If the acidified polysulphide filtrate is liver-colored or brown, copper 
is probably present, since CuS is somewhat soluble in (NH 4 ) 2 S j; . Con¬ 
siderable copper sulphide may be removed and recovered by digesting 
the residue with Na 2 S x in which the arsenic group sulphides are insolu¬ 
ble, but which dissolves very little CuS. The arsenic group may be 
precipitated from the solution as before. Na 2 S x could not be used to 
separate initially the copper and arsenic groups, as it dissolves HgS. 

ANALYSIS OF GROUP II-A 

32. Boil the precipitated copper group sulphides (residue 
from 31) with 5 c.c. HNO3 and 10 c.c. water for two minutes. 
Filter and wash the residue. 

The residue is black HgS. The residue is sometimes white, due to 
the formation of double salts of mercury with H 2 S and HN0 3 . 

Heat the residue with cone. HC1 and one or two crystals 
of KCIO 3 . Boil until all the chlorine is expelled, dilute with 
a little water and add SnCh. A white precipitate of Hg 2 Cl 2 
or black Hg is a test for mercuric mercury. 

33. Evaporate the filtrate from 32 with 5 c.c. cone. 
H2SO4 until white fumes of IT2SO4 escape, dilute to 20 c.c., 
allow to stand a few minutes, filter, and wash the lead sul¬ 
phate. Confirm the presence of lead as prescribed in Group 
I (20). 

The evaporation with H 2 S0 4 is necessary in order to expel excess 
HN0 3 in which PbS0 4 is quite soluble. Although the presence of lead 
may have been detected in Group I, any left in Group II must be 
removed, as it would interfere with later tests. What tests would be 
interfered with? 


64 


QUALITATIVE CHEMICAL ANALYSIS 


34. Make the filtrate from the PbSCL alkaline with 
NH4OH. A white precipitate of Bi(OTI) 3 is an indication of 
bismut 1 . Filter and pour over the residue Na2Sn02, freshly 
made by adding excess NaOH to SnCL. A black residue of 
bismuth is a confirmatory test for this base. 

In making up the sodium stannite solution, the beginner should 
test the reagent with litmus to make sure an actual excess of NaOH 
is present. 

35. If the filtrate from the Bi(OH )3 is blue, copper is 
present. Confirm this test or test for traces of the metal 
by acidifying a small portion of the solution with acetic 
acid and adding a little K 4 Fe(CN)6. A red or pink precip¬ 
itate of copper ferrocyanide is a test for copper. 

In the confirmatory test cadmium salts would give a white precipi¬ 
tate of cadmium ferrocyanide under the same conditions. 

36. If copper is present, add KCN in very slight excess 
of the amount necessary to discharge the blue color of the 
main part of the solution in 35. Pass in H 2 S; a yellow pre¬ 
cipitate of CdS is a test for cadmium. 

If the precipitate is black due to the presence of some black sulphide 
such as HgS or PbS, filter, and repeat the procedure beginning at 32. 


ANALYSIS OF GROUP II-B 

37. Method I. Heat the arsenic group sulphides (which, 
have been partially dried on the suction filter) with 10 c.c. 
cone. HC1 in a test tube immersed in boiling water for ten 
minutes. 1 

) N °y es and Bra y- J- Amer - Chem. Soc., 29 , 176 (1907). Their method for 
antimony and tin may be followed if desired, in the analysis of the filtrate 
diluted to 50 c.c. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 65 

As 2 S 6 is insoluble in cone. HC1 under these conditions whereas Sb 2 So 
and SnS 2 dissolve. 

If the H 2 S precipitate of Group II had not been treated with 
(NH 4 ) 2 S X (31-a), As 2 S 3 might be present in the arsenic group sulphides. 
It would be partially dissolved by cone. HC1 in the above process. 
Solution may be prevented by saturating the mixture containing the 
sulphides plus the cone. HC1 with H 2 S and passing the gas through 
the solution while it is being heated in the test tube. 

Add 5 c.c. water, filter, wash with 5 c.c. water adding the 
washings to the filtrate, and finally add 30 c.c. water to the 
filtrate so that its volume is 50 c.c. 

38. Wash the residue of As 2 S 5 (and As 2 S 3 ) from 37 with 
HC1 to remove antimony and tin, and heat with a little cone. 
HC1 and solid KCIO 3 . Boil until the chlorine is expelled and 
remove the sulphur. 

Free chlorine aids the solution of arsenic pentasulphide thus: 

As 2 S 5 + ioHC1 -> 2AsC1 5 +5H 2 S, 

H 2 S+C1 2 ->2HC1+S, 

2AsC1 5 +8H 2 0 -> 2H 3 As0 4 + ioHC1. 

To the solution add about one half its volume of NH 4 OH, 
filter if necessary (traces of mercury may have come through 
to this point) and add 1 c.c. magnesia mixture. If the pre¬ 
cipitate does not form immediately, stir, rubbing the sides 
of the beaker with the stirring-rod, and allow to stand in the 
cold for at least thirty minutes. A white, crystalline pre¬ 
cipitate of MgNFLiAsCU is an indication of arsenic. Filter, 
wash, and dissolve the residue in HC1. Pass H 2 S into the 
solution heated nearly to boiling, for ten minutes. A yellow 
precipitate of As 2 S 3 (and As 2 S 5 ) is a confirmatory test for 
arsenic. 


66 


QUALITATIVE CHEMICAL ANALYSIS 


39. Boil the filtrate from 37 containing antimony and tin 
chlorides until all the H 2 S is expelled. 

Why is this necessary? 

The boiling should be continued until filter paper moistened with lead 
acetate solution and held over the vessel shows no blackening. 

Divide the solution into two portions and to one add 
some pure iron wire. Heat until the iron has dissolved, 
filter, and wash. Test the filtrate for tin according to 40 
immediately. If there is a black residue (antimony or a 
trace of carbon) treat it with a solution of two or three grams 
tartaric acid in 10 c.c. water to which 0.5 c.c. HNO3 is added. 
Dilute to 25 c.c. and saturate with H 2 S. An orange precipi¬ 
tate of Sb 2 S 3 is a test for antimony. 

In the second portion of the solution place a piece of plati¬ 
num foil on which is a piece of pure tin. A black deposit 
of metal is a test for antimony. 

In this last test the antimony may be confirmed by converting into 
the sulphide as in the previous test. If copper has come through to 
this point it would be deposited on the platinum as a reddish stain which 
could not be confused with antimony. 

40. Dilute the filtrate (39) with an equal volume of water, 
and add HgCl 2 . A white precipitate of \Hg 2 Cl 2 is a test for 

tin. 

SnCl 4 is reduced to SnCl 2 by the action of iron and ll Cl in 39;—As 
SnCl 2 rapidly oxidizes in hot solution especially, the test for tin must 
be made immediately after reduction. 

ANALYSIS OF GROUP II-B 

41. Method II. Warm the precipitated arsenic group 
sulphides (which in this case do not need to be filtered by 
suction) with cone. HC1 and if more than a residue of sul- 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 67 


phur remains add a little solid KCIO 3 and dissolve AS 2 S 5 . 
Expel excess chlorine, dilute, and filter. Place the solution 
in the generator of Marsh’s apparatus. This consists of a 
stout Erlenmeyer flask fitted with a thistle tube and a 
delivery tube. From the delivery tube the escaping gases 
pass through a CaCE tube and then through a hard glass 
tube drawn out at two points to a capillary, and finally 
they are passed into a solution of AgNOs. In the generator is 
placed some metallic zinc, a piece of platinum foil, and dilute 
H2SO4. Allow hydrogen to be developed for a few minutes 
until a portion of it collected in a test tube shows the,absence 
of oxygen. Now add the solution of the arsenic group in 
small portions. Heat the hard glass tube just back of the 
first capillary and a mirror of metallic antimony will be 
deposited a little beyond the heated part of the tube; the 
more volatile arsenic will be deposited as a mirror further 
along the tube. In the generator the SbCl 3 and H 3 ASO 4 are 
reduced to stibine, SbH 3 , and arsine, AsH 3 , and these gases 
are partially decomposed into their elements when passed 
through the heated tube. Some of the antimony is deposited 
on the platinum foil in the generator which serves as an 
additional test for this element. The action of the AgN0 3 
results in a precipitation of silver by AsH 3 and of silver anti- 
monide by SbH3. 

6AgN0 3 -f AsH 3 + 3 H 2 0 -> 6Ag+H 3 As0 3 +6HN0 3 , 
SbH 3 +3AgN0 3 SbAg 3 +3HN0 3 . 

The solution is filtered and the residue of Ag and SbAg 3 
washed, dissolved in HC1: 

SbAg 3 +6HCl -> SbCl 3 +3AgCl+3H 2 , 


68 QUALITATIVE CHEMICAL ANALYSIS 

and filtered from the AgCl. Into the filtrate is passed H2S, 
and orange Sb2S3 confirms antimony. If some AgCl is not 
all precipitated it will interfere with the test for antimony and 
should be removed by a little Kl. 

To the filtrate from the Ag and SbAg3 containing H3ASO3 
add NaCl to remove excess AgNC>3, filter, and pass H2S into 
the filtrate. A yellow precipitate of AS2S3 confirms the pres¬ 
ence of arsenic. 

The hard glass tube is immersed in sodium hypochlorite 
solution. The antimony mirror will be unattacked while the 
arsenic mirror will dissolve: 

As 2 + 5Na0Cl+3H 2 0 —> 5NaCl + 2H3As04. 

The residue of spongy tin and zinc left in the generator 
is dissolved in hot cone. HC 1 , diluted and tested for tin with 
HgCl 2 as in ( 40 ). 

Group III. The Iron and Zinc Groups. Fe +3 , Al +3 , 
Cr+ 3 and Mn+ 2 , Zn+ 2 , Co+ 2 , Ni+ 2 

Reactions 

IRON 

42 . Iron usually has a valence of two or three. There 
are three oxides—white ferrous oxide FeO, brownish-red 
ferric oxide Fe 2 03, and black ferrous-ferric oxide Fe3C>4. 
To FeO and Fe 2 03 correspond the hydroxides Fe(OH) 2 and 
Fe(OH)3, and ferrous and ferric salts with the colorless 
ferrous ions Fe+ 2 and the brownish-yellow ferric ions Fe+ 3 
respectively. Both series of salts hydrolyze in water, the 
ferric more largely than the ferrous salts. Ferrous salts are 


REACTIONS OF THE B&SE-FORMING CONSTITUENTS 69 

very unstable and especially easily in solution oxidize to 
ferric salts: 

Fe +2 Fe +3 . 

Fe 304 may be considered to be a ferrous salt, FeFe 2 04 , of 
the hypothetical ferric acid, H 2 Fe 2 0 4 . Other ferrites such as 
ZnFe 2 04 are known. (Compare the amphoteric nature of 
hydroxides containing ferric iron with aluminium hydroxide.) 

1. Heat an iron compound in the microcosmic salt bead 
in the reducing and oxidizing flames. 

2 . Heat an iron compound with Na 2 C 03 on charcoal'. 
Test the black mass with a magnet. 

3 . Test the reaction of solutions of ferrous and ferric salts 
to litmus and explain. 

Ferrous salt solutions are made up directly before use on account of 
their instability. 

4 . Alkali hydroxides precipitate from solutions of ferrous 
salts Fe(OH) 2 , oxidizing rapidly in the air to Fe(OH) 3 . 
Note the changes in color. Na 2 C 03 precipitates FeC 03 , 
oxidizing rapidly to Fe(OH) 3 . 

5 . (NH 4 ) 2 S (or NH4OH plus H 2 S) precipitates FeS from 
ferrous salts. FeS is soluble in HC1. 

Calculate the solubility product for FeS and compare it with that of 
CuS. 

6 . KCN precipitates Fe(CN ) 2 from ferrous salts, soluble 
in excess, forming the salt of a complex ferrous ion, potas¬ 
sium ferrocyanide: 

4 KCN+Fe(CN ) 2 -> K 4 Fe(CN) 6 . 

By electrolysis the iron can be shown to be in the anion. The ion 
[Fe(CN) 6 ] -4 is very stable, as can be proved by adding KOH. No 


70 


QUALITATIVE CHEMICAL ANALYSIS 


precipitate is observed, showing the absence of an appreciable number 
of Fe +2 ions. The corresponding acid H 4 Fe(CN ) 4 is known. 

Ferrous salts produce with K 4 Fe(CN)o a white precipitate 
of K2FeFe(CN)6, potassium ferrous ferrocyanide, which oxi¬ 
dizes rapidly in the air forming Prussian blue (see below). 

6K2FeFe(CN) 6 + 3 H 2 0+30 -> Fe 4 [Fe(CN)o]3 

+ 3 K 4 Fe(CN ) 6 + 2Fe(OH) 3 . 

Ferric salts give immediately with K 4 Fe(CN )6 a precipitate of 
Prussian blue, Fe 4 [Fe(CN)e]3- This is a test for the Fe +3 
ion. 

K 4 Fe(CN)e is oxidized by chlorine to the stable potassium 
ferricyanide, K 3 Fe(CN)e: 

2K 4 Fe(CN ) 6 + Cl 2 -> 2K 3 Fe(CN ) 6 + 2KCl. 

With K 3 Fe(CN)o ferrous salts yield a precipitate of Turn- 
bull’s blue, Fe 3 [Fe(CN) 6 ]2 (test for Fe +2 ion). Both Prussian 
blue and Turnbull’s blue are insoluble in dilute acids, but are 
converted by NaOH into ferric hydroxide and sodium ferro¬ 
cyanide, and ferrous hydroxide and sodium ferricyanide re¬ 
spectively. 

7. Alkali hydroxides precipitate Fe(OH) 3 from ferric salts, 
insoluble in excess alkali. Na 2 C 0 3 precipitates a basic car¬ 
bonate which is completely hydrolyzed to Fe(OH) 3 on heat¬ 
ing. 

8. (NFLO2S precipitates Fe 2 S 3 from solutions of ferric 
salts. 

9. Pass H2S into FeCl 3 solution and note change of color 
and the precipitate. 

10 . KSCN or NH 4 SCN produces a blood-red coloration 
with ferric ions. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 71 


This color is due to undissociated ferric thiocyanate. The test should 
not be made in presence of much HN0 3 which itself produces a red color 
with thiocyanates. HgCl 2 also interferes with the test (destroys the red 
color probably by the formation of mercuric thiocyanate). 

11. Na2HP04 precipitates from ferric salts FePCU, insolu¬ 
ble in acetic acid, but soluble in mineral acids. 

The reactions of aluminium and chromium are analogous. 

12. Add sodium acetate solution to FeCl3 and observe the 
color. Dilute and boil the solution. Basic ferric acetate is 
precipitated: 

FeCl 3 + 3 NaC 2 H 3 0 2 ♦=* 3NaCl+Fe(C 2 H 3 0 2 ) 3 . 

Fe(C 2 H 3 0 2 ) 3 + 2H 2 0 <=> Fe( 0 H) 2 C 2 H 3 0 2 + 2HC 2 H 3 0 2 . 

Explain why excess sodium acetate would prevent the reversal of the 
latter reaction. 

13. Dissolve some tartaric acid in a solution of a ferric 
salt and add NH4OH. 

No precipitate is produced with excess tartaric acid and this is 
characteristic of ferric, aluminium, and chromium salts. Organic 
hydroxy-acids such as tartaric and citric acids, and other organic com¬ 
pounds as glycerol, cane sugar, etc., prevent many precipitation reactions 
by forming stable complexes with the metallic ions. These may be of 
the same nature as the copper complexes. Thus tartaric acid prevents 
the precipitation of Cu(OH) 2 by KOH by the formation of a complex 
copper-tartrate anion: 

H,C,H 1 0 6 + 2K0 H+Cu('0 H), -> K.+KC.H.CuO )- 2 + 4 H.O. 

ALUMINIUM 

43 . Aluminium has a valence of three, one oxide A 1 2 0 3 
with the corresponding hydroxide Al(OH) 3 , and salts with the 
colorless ion A 1 + 3 . The existence of salts with the alumin- 
ate ion A 10 2 _1 and the amphoteric nature of aluminium 


72 


QUALITATIVE CHEMICAL ANALYSIS 


hydroxide will be considered below. All aluminium salts and 
aluminates are largely hydrolyzed in aqueous solution. 

1. Heat aluminium sulphate with a very little Co(N0s)2 
on charcoal. 

The blue color is due to the formation of cobalt aluminate 
Co(A10 2 )2. Phosphates will give the same color with the cobalt salt. 
If the cobalt nitrate is in excess, the blue color of the aluminate will 
be hidden by black cobalt oxide. 

2. NH 4 OH precipitates Al(OH)3 from solutions of alu¬ 
minium salts, insoluble in slight excess of the reagent. 

3 . Add NaOH gradually to a solution of an aluminium 
salt until Al(OH)3 is precipitated. Test the solubility of the 
precipitate in HC 1 and in excess NaOH. In the latter case 
sodium aluminate, NaA102, is formed. 

Explain fully the amphoteric nature of Al(OH) 3 as shewn by these 
reactions, and explain its insolubility in excess NH 4 OH. (A large excess 
of NH 4 OH dissolves Al(OH) 3 to some extent.) 

4. Na2C03 or (NH^S precipitates Al(OH)3. 

Explain. 

. 5 . Test with Na 2 HP 04 and with sodium acetate. 

The results are analogous to those obtained with ferric iron. 

CHROMIUM 

44 . Chromium usually has a valence of two, three, or six. 
(Compounds are known with a higher valence.) The com¬ 
mon oxides are green chromic oxide, Cr203, and brownish- 
red chromium trioxide, Cr03-Cr 2 03 is the anhydride of the 
base Cr(OH)3, showing slight acid tendency, while Cr03 
is the anhydride of the fairly strong chromic acid, H2Cr04 
(which has not been isolated). Chromous oxide, CrO, is 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 73 

unknown, although its hydroxide, Cr(OH) 2 , and salts (which 
are very unstable) with the chromous ion, Cr+ 2 , are known. 
Salts of Cr(OH)3 have the green chromic ion Cr +3 and the 
green chromite ion Cr 0 2 ~ 1 . Salts of chromic acid have the 
yellow chromate ion Cr 0 4 " 2 or the orange-red bichromate 
ion Cr 2 07 -2 . Salts of Cr(OH) 3 are hydrolyzed in water. 

1. Heat a chromium compound in the microcosmic salt 
bead in the reducing and oxidizing flames. 

2. Heat chromium nitrate with Na 2 C 0 3 and a little KN 0 3 
to fusion in a porcelain crucible. 

Explain this case of oxidation to a chromate. 

3. NH4OH, (NH 4 ) 2 S, or Na 2 C 0 3 , precipitates Cr(OH) 3 
from solutions of chromium salts. 

Cr(OH) 3 is slightly soluble in excess cold NH 4 OH, possibly due to the 
formation of chromium-ammonium compounds. It is completely pre¬ 
cipitated by boiling the solution. 

4. NaOH precipitates Cr(OH) 3 from chromium salts, 
dissolving in excess with the formation of sodium chromite, 
NaCr 0 2 . 

When a solution of sodium chromite is boiled it is hydrolyzed and 
chromium is precipitated as a completely insoluble hydrate (variable 
composition). Compare this action with that of aluminium. 

5. Na 2 HP 0 4 precipitates CrP 0 4 from chromium salts. 

Sodium acetate does not yield a precipitate with chromium salts 
(compare iron and aluminium). 

6 . Add excess NaOH to a chromium salt and then sodium 
peroxide, Na 2 0 2 . (Do not touch Na 2 0 2 with the hands, as 
it may cause dangerous wounds. It ignites paper and other 
organic matter, and may be transferred by use of a porce¬ 
lain spatula.) Repeat the experiment, using bromine instead 
of sodium peroxide. 


74 


QUALITATIVE CHEMICAL ANALYSIS 


The following will indicate the method of oxidation of sodium chro¬ 
mite by sodium peroxide: 


Na 2 0 2 +H 2 0 2 NaOH+ 0 , 

2NaCr0 2 + 2Na0H+30 -* 2Na 2 Cr0 4 TH 2 0. 

7. Pass H2S into a solution of K2Cr04 acidified with 
HC 1 . 

When HC 1 is added to K 2 Cr 0 4 the following reaction takes place: 
K 2 Cr 0 4 +HCl -» KHCr 0 4 +KCl. 

KHCr 0 4 is unstable, however, and loses water: 

2 KHCr 0 4 -> K 2 Cr 2 0 7 +H 2 0 , 

potassium dichromate being formed. This salt, therefore, is the anhy¬ 
dride of potassium acid chromate. Now when H 2 S is added, the fol¬ 
lowing reactions may be assumed to take place in the direction indi¬ 
cated, largely due to the fact that free sulphur is formed: 

K 2 Cr 2 0 7 +2HCI -> H 2 Cr 2 0 7 +2KCl; 

the acid formed will have some tendency to ionize as a base (the 
hydroxides of chromium are amphoteric and in this case the acid 
dissociation will be suppressed by the presence of HC 1 ): 

H 2 Cr 2 0 7 + 5 H 2 0 -> 2Cr(OH) 6 , 


and the base will form a salt with HC 1 : 

Cr(OH)«+6HCl CrCl 6 + 3 II 2 0 , 
which will be reduced by the H 2 S: 

2CrCl 6 +3H 2 S 2CrCl 3 + 6 HC 1 +3S. 

Express these reactions in the ionic form. 1 

1 See Stieglitz, Qual. Anal., 11, p. 288. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 75 


MANGANESE 

45 . Manganese may bave a valence of two, three, four, 
six, or seven in its compounds. Green manganous oxide, 
MnO, anhydride of the base Mn(OH)2, is the only oxide 
yielding very stable salts. These have the pink manganous 
ion Mn +2 , and those salts having anions of strong acids (HC 1 ) 
are only little hydrolyzed. Brownish-black manganic oxide, 
Mn 2 03, in some ways acts as if it were manganese manganite, 
MnMn 0 3 , but compounds of trivalent manganese such as 
Mn2(S04)3 and the hydroxide Mn(OH) 3 are known. Red 
trimanganese tetroxide, Mn 3 0 4 , may be considered to be a 
manganese compound (M^MnCU) of orthomanganous acid, 
Mn(OH) 4 , since with nitric acid Mn 3 0 4 yields manganese 
nitrate and orthomanganous acid (which latter loses water, 
forming metamanganous acid, MnO(OH) 2 ). Black manganese 
dioxide, Mn 0 2 , is the anhydride of the acid MnO(OH) 2 
(or H 2 Mn 0 3 ) corresponding to the manganites such as 
CaMnOs. Mn 0 2 also has an unstable hydrate Mn(OH) 4 
with some basic tendency, since the existence of MnCl 4 has 
been established. Red manganese trioxide, MnC>3, is very 
unstable and is the anhydride of manganic acid, H2Mn0 4 , 
(which has not been isolated), which yields the manganates 
with the unstable green ions Mn 0 4 ~ 2 . Manganese heptox- 
ide, Mn 2 C>7, is a^ brownish oil. It is the anhydride of per¬ 
manganic acid, Mn 0 3 ( 0 H) (or HMn 0 4 ), whose salts yield the 
purple ion Mn 0 4 _1 . 

1. Heat some MnS 0 4 in the microcosmic salt bead in the 
reducing and oxidizing flames. 

2. Fuse a manganese compound with Na 2 C 0 3 and a little 
KNO3 on a porcelain crucible cover or a piece of platinum foil. 


76 


QUALITATIVE CHEMICAL ANALYSIS 


Sodium manganate, Na^MnCh, is formed. 

3. NH4OH precipitates from solutions of manganous salts 
Mn(OH) 2 , readily soluble in ammonium salts. 

Explain the effect of the ammonium salts. 

4. NaOH precipitates Mn(OH) 2 , from manganous salts. 
Test the solubility of the hydroxide in excess NaOH. 

Mn(OH) 2 changes color on standing, duei to oxidation. Manganous 
acid is first formed: 

Mn( 0 H ) 2 +0 -> MnO(OH) 2 , 

and this forms a salt (manganous manganite) with the unchanged 
hydroxide: 

Mn(OH) 2 +H 2 MnOd —*MnMn0 3 +2H 2 0. 

Write these reactions in the ionic form. 

Add Na 2 0 2 and bromine to separate portions of the alka¬ 
line mixture. 

Manganous acid is formed in both cases. Bromine forms sodium 
hypobromite with NaOH and this salt acts as follows: 

NaBrO —* NaBr+O. 

Write all reactions with full explanations. 

5. (NH) 2 S precipitates MnS from manganous salts. MnS 
is soluble in HC1. 

6. Na 2 C03 precipitates MnCOs. 

7. Boil a solution of Mn(NOs)2 with cone. HNO3 and a 
little solid KCIO3. 

Mn 0 2 is precipitated and C 10 2 evolved. As Mn 0 2 is soluble in HC 1 , 
no chlorides should be present. HC 1 may be expelled from such solu¬ 
tions by continued evaporation with HN 0 3 . 

8. Add to cone. HNO3 some lead dioxide, Pb 0 2 , and man¬ 
ganous sulphate, MnSC>4, boil, and allow to settle. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 77 

The purple color is due to the formation of permanganic acid, 
HMn0 4 \ 


ZINC 

46 . Zinc always has a valence of two. White zinc oxide, 
ZnO, yields an hydroxide, Zn(0H)2, which is amphoteric. 
Salts of zinc hydroxide have either the colorless zinc ion 
Zn+ 2 or the colorless zincate ion Zn 0 2 “ 2 . 

1. Heat a zinc compound with Na 2 C 0 3 on charcoal and 
note the color of the incrustation while hot and cold. Add 
a trace of cobalt nitrate and heat again. 

2. NH4OH precipitates Zn(OH) 2 from solutions of zinc 
salts. 

The precipitate is soluble in excess ammonium hydroxide for the 
same reasons that silver and copper hydroxides so dissolve. How 
would the stability of the zinc-ammonium ion Zn(NH 0 ) 4 + 2 be affected by 
a small and a large excess of ammonia? The precipitate of zinc hydrox¬ 
ide is also soluble in ammonium salts, which by the common ion effect 
decrease the amount of hydroxyl ions and increase the amount of NH a : 

NH 4 +1 +OH _1 NH 4 OH NH a +H 2 0 . 

Explain more fully the action of the ammonium ions. 

3. NaOH precipitates Zn(OH) 2 from zinc salts, readily 
soluble in HC 1 or excess NaOH. 

Sodium zincate, Na 2 Zn 0 2 , is formed in the latter case. Write these 
reactions in the ionic form. 

4. (NH 4 )2S precipitates ZnS, insoluble in acetic acid, solu¬ 
ble in hydrochloric acid. 

Explain these reactions by use of the solubility-product principle. 

5. Na 2 C 0 3 precipitates a basic carbonate. 


78 


QUALITATIVE CHEMICAL ANALYSIS 


COBALT 

47 . The usual valence of cobalt in its compounds is two 
or three. Black cobaltous oxide, CoO, is the anhydride of 
the base Co (OH) 2 which shows only slight amphoteric prop¬ 
erties; its salts yield the pinkish-red ion C0+ 2 . Warm 
concentrated solutions of cobaltous chloride or solutions to 
which an excess of HC 1 is added are blue. This may be 
due to the suppression of the ionization of the salt by con¬ 
centration or the effect of the common ion of HC 1 . (Anhy¬ 
drous cobaltous salts are blue.) 

Co+ 2 + 2 C 1 - 1 ^CoC 1 2 . 

Cobaltic oxide, C02O3, is the anhydride of the base Co(OH) 3 , 
which is weaker than cobaltous hydroxide, cobaltic salts 
being completely hydrolyzed in solution. Cobalto-cobaltic 
oxide, C03O4, may be considered to be a cobaltous compound 
of hypothetical cobaltic acid, H2C02O4. 

1. Try the microcosmic salt bead test with a cobalt com¬ 
pound. 

2. Heat a cobalt compound with Na 2 CC>3 on charcoal. 
Test the metal with a magnet. 

Use a solution of a cobaltous salt for the following tests. 

3. NH4OH precipitates a basic salt of variable compo¬ 
sition (e.g., Co(OH)Cl) which readily dissolves in excess, 
forming complex cobalt-ammonium compounds, of which 
Co(NH 3 )4C1 is an example. 

From observations in the next experiment, why is cobalt supposed 
to be in the cation and not in the anion of this salt, as it would be if 
cobaltous hydroxide dissolved in NH 4 OH on account of amphoterisis? 

4. NaOH precipitates a basic salt which is converted 
into Co(OH) 2 by heating. (Note the change of color.) Only 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 79 


an extremely concentrated solution of NaOH will dissolve 
cobaltous hydroxide. 

Bromine or sodium peroxide oxidizes cobaltous hydroxide 
to cobaltic hydroxide, Co(OH) 3 . 

Add to the neutral solution of a cobaltous salt KCN, drop by 
drop, until the precipitate of Co(CN) 2 first formed dissolves. The 
potassium cobaltocyanide which results will yield a precipitate with 
sodium hypobromite because of the instability of the complex cobalto¬ 
cyanide ion: 

(Co(CN f )]-« *=± Co +2 -f6CN _1 . 

If the solution of the potassium cobaltocyanide is warmed for some 
time, it will not yield a precipitate of Co(OH) 3 with sodium hypo¬ 
bromite. It is oxidized in the air: 

2 K 4 Co (CN) e+O -f H 2 0 — 2K 3 Co(CN) 6 +2KOH, 

and the stable cobalticyanide ion yields insufficient cobaltic ions for 
cobaltic hydroxide to be precipitated. 

[Co(CN) 6 J- 3 <=± Co +3 +6CN -1 . 

5. (NH^S precipitates CoS. 

CoS is only very slowly dissolved by cold, very dilute HC 1 . Since 
H 2 S causes no precipitation in solutions of cobalt chloride acidified with 
HC 1 , the insolubility of cobaltous sulphide in the acid must be due to a 
very slow reaction velocity towards the right in the following: * 

CoS+ 2 HC 1 ^±CoC 1 2 +H 2 S, 

and therefore much time is required for the true equilibrium to be 
reached. 

CoS is readily dissolved by aqua regia , or by cone. HC 1 plus KC 10 3 . 

6. Na2C0 3 precipitates a basic salt. 

7. KNO2 precipitates potassium cobaltic nitrite in the 
presence of acetic acid, and more readily if the solution is 
warmed. 


80 


QUALITATIVE CHEMICAL ANALYSIS 


Cobaltous nitrite is first formed and this is oxidized by nitrous acid, 
liberated from the potassium nitrite, to cobaltic nitrite: 

Co(N 0 2 ) 2 + 2HN0 2 -> Co(N 0 2 ) 3 +H 2 0 +NO, 

and the cobaltic salt combines with potassium nitrite to form a salt with 
a complex cobaltic anion: 

Co(N 0 2 ) 3 + 3 KN 0 2 -+K 3 Co(N 0 2 ) 6 . 

8 . To a solution of cobalt sulphate add an equal volume 
of glacial acetic acid, and then add a few drops of nitroso- 
beta-naphthol, HNO2C10H6. 

The precipitate is a cobalt salt of the naphthol, Co(NO 2 Ci 0 H 6 ).;. 
Glacial acetic acid (100 per cent HC 2 H 3 0 2 ) is added to the test solution 
since the reagent is made by dissolving nitroso-beta-naphthol in 50 per 
cent acetic acid and is precipitated itself on dilution. Nitric acid inter¬ 
feres with the test by action on the reagent, so that cobalt sulphate 
is used. 

Nickel is not precipitated by this reagent. 

NICKEL 

48 . Nickel usually has a valence of two or three in its 
compounds. Green nickelous oxide, NiO, is the anhydride 
of the base Ni(OH)2, whose salts have the green nickelous 
ion Ni+ 2 . Black nickelic oxide, Ni203, is the anhydride of 
the hydroxide Ni(OH) 3 , which is not basic and which yields 
no salts. Nis04 is also known. Nickel forms complex salts 
less stable than the analogous cobalt compounds such as the 
cobaltocyanides; as nickelic salts are unknown, there are no 
such salts as those with a nickelicyanide ion. 

1. Test a nickelic compound in the microcosmic salt bead. 

2. Heat a nickel compound on charcoal with Na2C03. 
Test the magnetic properties of the metal. 

Use a solution of a nickel salt for the following tests. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 81 

3 * NH 4 OH precipitates a basic salt, dissolving in excess 
with the formation of nickel-ammonium ions, [Ni(NH 3 ) 4 ]+ 2 . 

Compare the color with that of the corresponding copper-ammnoium 
10ns. 

NH4OH produces no precipitate in solutions of nickel salts to which 
ammonium salts are added. (Explain). 

4. NaOH precipitates Ni(OH) 2 . 

Bromine and Na 2 0 2 act on Ni(OH) 2 in the same way in 
which they act on cobaltous hydroxide. 

KCN precipitates Ni(CN) 2 from neutral solutions of nickel 
salts, and the precipitate dissolves in excess of the reagent, 
forming potassium nickelocyanide, K 2 Ni(CN) 4 . The com' 
plex nickelocyanide ions yield enough nickel ions: 

[Ni(CN) 4 ]~ 2 <=> Ni +2 + 4 CN~ 1 , 

for bromine to precipitate Ni(OH) 3 from such solutions. 

5. (NH 4 ) 2 S precipitates NiS. Nickel sulphide dissolves 
to some extent in excess ammonium sulphide, and if filtered, 
a brown filtrate is obtained. This solution, which holds the 
nickel either in the form of a colloidal sulphide or as a poly¬ 
sulphide, may be coagulated by either boiling with a little 
acetic acid or adding some ammonium chloride solution and 
passing in H 2 S. 

6. Na 2 C 0 3 precipitates NiC 0 3 

7. Note that in dilute solutions, KN 0 2 produces no pre¬ 
cipitate with nickel salts, but if these are very concentrated 
and the solution is warm, a brownish-red precipitate of potas¬ 
sium nickelonitrite, K 4 Ni(N 0 2 ) 6 , will be formed. 

8. Dime thy 1 -gly oxime, H 2 N 2 0 2 C 4 H 6 , precipitates the nickel 
salt NiN 2 0 2 C 4 H G from solutions of nickel salts slightly alka¬ 
line with NH 4 OH. 


82 


QUALITATIVE CHEMICAL ANALYSIS 


The nickel salt of dimethyl-glyoxime is also insoluble in solutions 
slightly acid with acetic acid. 

Cobalt is not precipitated by this reagent. 


ANALYSIS OF GROUP III 

49 . Precipitation of the Iron Group. Boil the filtrate 

from the H2S precipitate of Group II until the H2S is all 
expelled (test with lead acetate paper). Add 2 c.c. bromine 
water and expel excess bromine by heating. 

The H 2 S is removed so that the indications furnished by the action 
of NH4OH will not be obscured by the precipitation of sulphides of 
Group III. Bromine oxidizes the iron to the ferric condition. 

Add 15 c.c. NH4CI, set aside 5 c.c. of the solution, and 
make the larger portion faintly alkaline by adding NH4OH 
in small amounts until after shaking the odor of ammonia 
persists. Then add 2 c.c. in excess. 

The added NH 4 C 1 , together with the ammonium salts formed by 
neutralization of the acid solution, lessen the solubility of Al(OH) 3 and 
also prevent the precipitation of Mg(OH) 2 and Mn(OH) 2 . Why? 

A large excess of NH 4 OH is avoided, since NiS, as formed below, has 
less tendency to run through the filter in solutions only faintly ammo- 
niacal. 

If there is no precipitate, make the 5 c.c. of reserved solu¬ 
tion alkaline with NH4OH and combine with the remainder 
of the solution. 

If arsenic has been found present, the solution may be advantage¬ 
ously tested for H 3 P 0 4 at this point, whether NH 4 OH forms a precipitate 
or not, since arsenic must be removed from a solution in which ammo¬ 
nium molybdate is to serve as a reagent for H 3 P 0 4 . 

If there is a precipitate, test the 5 c.c. of reserved solu¬ 
tion for phosphoric acid by pouring it into three times its 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 83 

volume of ammonium molybdate solution. Warm gently; a 
yellow precipitate of ammonium phosphomolybdate, 

(NH 4 ) 3 P 0 4 • 12M0O3 • 6 H 2 0 , 

is a test for phosphoric acid. If phosphoric acid is present, 
proceed with the analysis with the modifications outlined 
below. 

In the absence of phosphoric acid, NH 4 OH precipitates only red 
Fe(OH) 3 , grayish-green Cr(OH) 3 , and white Al(OH) 3 . If no precipi¬ 
tate is formed, iron, chromium, and aluminium are absent (as base¬ 
forming constituents). 

If phosphoric acid is present, the NH 4 OH precipitate may also con¬ 
sist of manganese and alkaline earth phosphates, since these salts are 
insoluble in an ammoniacal solution. Since phosphates of the trivalent 
metals are more insoluble than those of the bivalent metals, if sufficient 
iron, aluminium, or chromium is present, none of the bivalent metal 
phosphates will be precipitated. 

Oxalates, borates, and fluorides of the alkaline earths are also more 
or less insoluble in NH 4 OH. Hydrofluoric acid will have been removed 
in the preparation of the solution of alkaline earth fluorides. Further, 
the fluorides and borates are largely soluble in excess of ammonium salts, 
so that their presence causes no appreciable trouble in the ordinary 
analysis. Oxalic acid may be removed from the original solution 
according to 113 if its presence is found to cause serious trouble in the 
analysis. 

50. Precipitation of the Zinc Group. Saturate the am¬ 
moniacal solution of 49 (without filtering) with H2S; boil, 
filter, and wash with water containing a little ammonium 
sulphide and ammonium chloride. Reserve the filtrate for 
the analysis of Groups IV and V. 

By this procedure, the metals of the third group are all precipitated 
together—(NH 4 ) 2 S (H 2 S plus NH 4 OH) being the group precipitant. 
Al(OH) 3 and Cr(OH) 3 are unchanged by H 2 S; Fe(OH) 3 is converted into 
black Fe 2 S 3 ; white ZnS, flesh-colored MnS, and black CoS and NiS 
are precipitated. 


84 QUALITATIVE CHEMICAL ANALYSIS 

The ammonium sulphide wash water prevents the oxidation of the 
sulphides to soluble sulphates and the ammonium chloride hinders 
colloid formation. 

If the (NH 4 ) 2 S filtrate is brown, nickel sulphide has passed through 
the filter. In such a case, make the filtrate slightly acid with acetic 
acid, boil for several minutes, and filter. If the filtrate is still brown, 
boil again, and repeat until a clear filtrate is obtained. Add the residue 
of NiS to the residue containing the entire Group III metals. 

The iron group is not separated from the zinc group by filtering 
the NH 4 OH solution, since the separation is not always complete. In 
view of the amphoteric nature of some of the metals, with certain 
combinations, such salts as zinc chromite, ZnCr 2 0 4 , may be precipitated. 

51 . Isolation of CoS and NiS. To the (NH^S residue 
from 50 , transferred to a beaker, add 50 to 100 c.c. cold 
dilute HC 1 , made by diluting the 6-normal reagent with four 
volumes of water. If there is a residue left after standing 
one minute, filter as rapidly as possible to prevent con¬ 
tinued solution of CoS and NiS. Reserve the filtrate for 55 . 

52 . Dissolve the residue from 51 in 5 to 10 c.c. cone. HC 1 
to which a crystal of KCIO3 is added. Evaporate the solu¬ 
tion almost to dryness and dilute with 10 c.c. water. 

53 . Detection of Cobalt and Nickel. Method I. Make 
the diluted solution of cobalt and nickel chlorides (freed 
largely from acid by evaporation ( 52 )) barely alkaline with 
NH4OH and filter (from Fe(OH) 3 ) if necessary. Divide the 
filtrate into two equal portions. 

To one portion add an equal volume of glacial acetic 
acid. After this, add 2 c.c. nitroso-beta-naphthol and heat 
to boiling. A brick-red precipitate of Co(NO2Ci 0 H 6 )3 is a 
test for cobalt. Filter, wash free from nickel with 50 per 
cent acetic acid, and confirm the presence of cobalt in the 
residue by the metaphosphate bead test. 

To the second portion of solution add 2 c.c. dimethyl- 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 85 

glyoxime and heat to boiling. A pinkish-red precipitate of 
N1N2O2C4H6 is a test for nickel. Filter, wash, and confirm 
nickel by the metaphosphate bead test. 

54. Detection of Cobalt and Nickel. Method II. Add 
NaOH to the solution (52) until it is neutral or slightly 
alkaline. Divide the solution into two equal portions. 

Make one portion acid with acetic acid, adding 2 c.c. in 
excess, and then add to the solution nearly three times its 
volume of KNO2. If a precipitate does not immediately 
form, set aside for an hour, preferably in a warm place. A 
yellow precipitate of potassium cobaltinitrite is a test for 
cobalt. Confirm its presence by filtering, washing, and testing 
the residue in the metaphosphate bead. 

To the second portion of the solution, add KCN gradually 
until the colored cyanides dissolve, and then add a few drops 
of KCN in excess. Warm the solution for a few minutes to 
oxidize the cobalt salt, filter if necessary, and then add NaOH 
and an excess of bromine water. A black precipitate of 
Ni(OH)3 is a test for nickel. Filter, wash thoroughly, and 
confirm the presence of nickel by the bead test. 

An excess of bromine is present when some of the solution (acidified) 
is found to color starch-iodide paper blue. 

55. Separation of Aluminium, Chromium, and Zinc, from Iron, 
Manganese (and alkaline earths). Evaporate the filtrate 
from 51 nearly to dryness, add NaOH until alkaline, and then 
to the cold solution add 3 to 4 grams Na202. If H3PO4 has 
been found to be present, add 3 to 4 grams Na2C03. Heat 
until all the Na202 is decomposed as evidenced by cessation 
of the evolution of oxygen, dilute to 50 c.c., and filter. 

Na 2 C 0 3 is added if phosphate is present, to precipitate, as the more 
insoluble carbonates, any alkaline earth hydroxides which might be 


86 QUALITATIVE CHEMICAL ANALYSIS 

formed and which are somewhat soluble in the strongly alkaline solu- 

tl0 NaOH precipitates red Fe(OH),, white Mn(OH) 2 , and traces of green 
Ni(OH)j and blue basic salts of cobalt. Some zinc may be present in 
the precipitate, especially in the presence of large amounts of man- 
ganese; therefore tests are made for it in both precipitate and nitrate. 
Sodium aluminate, sodium chromite, and sodium zincate are left in 
solution. NaOH may also precipitate phosphates of Groups III and IV. 

NaiOi oxidizes Mn(OH), to brown &MnO s , cobalt and nickel com¬ 
pounds to the black higher hydroxides, and sodium chromite to sodium 
chromate. 

56 . Analysis of the Na 2 0 2 Filtrate. Acidify the filtrate from 
55 with KN 0 3 (test with litmus), make alkaline with NH 4 OH, 
and filter. A white gelatinous precipitate of Al(OH) 3 is a 
test for aluminium. Confirm by igniting the residue on 
platinum foil with Co(N 0 3 ) 2 and observing the blue color. 

Explain the reasons for making the solution (alkaline with NaOH) 
first acid and then ammoniacal. 

If there is any doubt about the identity of the precipitate, and 
to further distinguish it fiom silicic acid, H 2 Si 0 3 , which is of exactly 
the same appearance, dissolve the residue in a little HC 1 , evaporate to 
dryness (do not ignite) and heat for a half an hour at 125 in the hot 
closet. Al(OH), is converted into A 1 2 0 3 , and silicic acid into Si 0 2 (or 
insoluble hydrate). Heat with a little HC 1 , dilute, and filter. The silica 
is insoluble in HC 1 . Make the filtrate alkaline with NH 4 OH. Al(OH) 3 
is precipitated if aluminium is present. 

57 . If the Na 2 0 2 filtrate is yellow, chromium is present. 
Make the following confirmatory test. Divide the filtrate 
from 56 into two portions. Make one portion faintly acid 
with acetic acid and add Pb(C 2 H 3 0 2 ) 2 . A yellow precipitate 
of PbCr 0 4 confirms the presence of chromium. 

58 . Add BaCl 2 to the second portion of the NH 4 OH 
filtrate from 56 . Filter from BaCr 0 4 (lemon-yellow) and 

1 Read Noyes, Bray, and Spear. J. Amer. Chem. Soc., 30, 501 (1908). 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 87 

pass in II 2 S. A white precipitate of ZnS is a test for zinc. 
Confirm by igniting on platinum foil with Co(N 0 3 ) 2 and 
observing the green color. 

White BaS 0 4 may be precipitated by BaCl 2 due to the presence of 
sulphate ion introduced into the solution through oxidation of the 
original (NH 4 ) 2 S precipitate. 

The test for chromium is made with lead acetate since lead chromate 
is more highly colored than barium chromate. 

Zinc may not be tested for in the solution to which lead acetate 
has been added, since black PbS would be precipitated. 

If chromium is absent, BaCl 2 need not be added before testing for 
zinc. The chromate, if present, is removed, as it would be slowly re¬ 
duced by H 2 S and give a precipitate of sulphur, which would obscure 
the zinc test. 

59. Analysis of the Na 2 0 2 Residue. I. Phosphoric Acid 
Absent. Dissolve the Na 2 0 2 residue from 55 in 5 c.c. cone. 
HC 1 , add 5 c.c. NH 4 C 1 , and then make the solution alkaline 
with NH4OH. A brownish-red precipitate of Fe(OH) 3 is a 
test for iron. Filter and reserve the filtrate for 60. Dissolve 
the residue in HC 1 and add KSCN. A blood-red color is a 
confirmatory test for iron. Test the original solution of the 
unknown with KSCN and with K 3 Fe(CN) 6 to determine 
the state of oxidation of the iron in that solution. 

The added NH 4 C 1 together with that formed by neutralization of the 
HC 1 prevents the precipitation of Mn(OH) 2 by NH 4 OH. 

60 . Acidify the filtrate from the Fe(OH) 3 with HC 1 , 
evaporate almost to dryness, and then make the solution 
alkaline with NaOH. Add 2 grams Na 2 0 2 , heat, dilute to 
25 c.c. and filter. 

The precipitate consists of brown H 2 Mn 0 3 and possible traces of 
Co(OH) 3 and Ni(OH) 3 . CoS and NiS are not completely insoluble in 
the dilute HC 1 used in 51, so that they may appear at this point in the 
analysis. However, not more than one milligram of these metals may 


88 


QUALITATIVE CHEMICAL ANALYSIS 


be present if the procedure in 51 has been properly carried out, so that 
the formation of a relatively larger amount of brown residue is indicative 
of the presence of manganese. 

Transfer a small portion of the residue to a porcelain 
dish, add io c.c. HNO3 and one gram of Pb02, and boil 
for at least two minutes. Pour the mixture into a test-tube 
and allow the Pb02 to settle. The distinctly purple color 
of HMnCL confirms the presence of manganese. 

The presence of one milligram of cobalt or nickel will not interfere 
with this test. 

If cobalt and nickel have not been previously detected, 
dissolve the remainder of the residue in 5 c.c. HC1, dilute to 
10 c.c. with water, and test for these bases according to 53 . 

The presence of manganese interferes in no way with the tests for 
cobalt and nickel with nitroso-beta-naphthol and dimethyl-glyoxime 
respectively. 

61 . Pass H2S into the filtrate from 60 . A white precipi¬ 
tate of ZnS is a test for zinc which may be carried down 
in the Na202 precipitate of 55 . Confirm by igniting with 
Co(NOs) 2 and observing the green color. 

62. Analysis of the Na 2 C >2 Residue. II. Phosphoric Acid 
Present. Dissolve the residue from 55 in 5 c.c. cone. HC 1 , 
dilute to 10 c.c. with water, and add a few drops of K4Fe(CN)6 
to 1 c.c. of the solution. A blue precipitate of Fe4[Fe(CN)c]3 
is a test for iron. Test the original solution of the unknown 
with KSCN and with K 3 Fe(CN)o to determine the state of 
oxidation of the iron in that solution. 

A test for iron is made in a portion of the solution at this point, 
since a ferric salt is to be added to the solution later. 

63 . Basic Acetate Separation of Iron and Phosphate from the 
Bivalent Metals. Evaporate the remainder of the HC1 solu- 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 89 

tion from 62 to a small bulk, and then add NH 4 OH, little 
by little, until the solution is neutral. If a permanent pre¬ 
cipitate forms, make the solution faintly acid with acetic 
acid. Add an equal bulk of ammonium acetate solution. If 
the solution is not now distinctly reddish in color, add FeCl 3 
until it becomes so. Dilute with an equal volume of water, 
boil several minutes, and filter. 

By this procedure, a satisfactory separation of phosphoric acid as 
FeP0 4 from manganese, cobalt, nickel, zinc, and the alkaline earths, 
can be made. 

Iron is added until there is sufficient present to combine with all 
the phosphoric acid, and FeP0 4 is precipitated, as it is more insoluble 
than the bivalent metal phosphates. 

Enough iron is present to unite with all the phosphate when the 
red color of ferric acetate is evidenced. Before this point is reached, 
yellow-white ferric phosphate precipitates. 

When the solution is boiled, the excess of iron is precipitated as a 
basic acetate by hydrolysis. 

The precipitate of iron salts may be rejected. 

64. Make the filtrate from 63 alkaline with NH 4 OH and 
saturate the solution with H 2 S. Filter and wash. Add the 
filtrate to the (NFL^S filtrate from 50 and analyze for 
Groups IV and V. 

% 

The filtrate from 63 contains salts of manganese (traces of cobalt, 
nickel, and zinc), calcium, barium, strontium, and magnesium. NH 4 OH 
and H 2 S precipitate all but the alkaline earths. 

It should be noted that in the absence of an excess of phosphoric 
acid the alkaline earths would be precipitated, only in part or not at 
all by NH 4 OH when this reagent is added to the original solution of 
Group III. 

Dissolve the (NELO 2 S residue in 5 c.c. cone. HC1 to which 
is added a crystal of KCIO 3 , evaporate almost to dryness, and 
test the solution for manganese, cobalt, nickel, and zinc, 
according to 60 and 61. 


90 


QUALITATIVE CHEMICAL ANALYSIS 


Group IV. The Alkaline Earth Group . 1 

Ba+ 2 , Sr+ 2 , Ca+ 2 , Mg+ 2 

65 . The metals of Group IV all have a valence of two in 
their compounds, with white oxides which yield more or less 
soluble basic hydroxides; barium hydroxide is the strongest 
and magnesium hydroxide is the weakest base. Saturated 
solutions of barium, strontium, and calcium hydroxides pre¬ 
cipitate metals of the first three groups and also magnesium, 
as hydroxides. Sodium hydroxide precipitates hydroxides of 
the alkaline earths to the extent of their insolubility in water. 


SOLUBILITY PRODUCTS OF ALKALINE EARTH SALTS 


Salts 

Temp. 

Ionic Product 

Solubility Prod¬ 
uct 

BaC 0 3 . 

16 

[Ba+ 2 ]X[C<V 2 ] 

1.9X10“ 9 

BaC 2 0 4 . 

18 

[Ba+ 2 ]X[C 2 0 4 - 2 ] 

1.2X10 “' 1 

BaCr 0 4 . 

18 

[Ba +2 iX[Cr 0 4 -2 ] 

1.1X io“ 10 

BaF 2 . 

18 

[Ba+ 2 ]X[F->] 2 

1.1X io -6 

BaS 0 4 . 

18 

[Ba+ 2 ]X[S 0 4 - 2 ] 

1. oX 10 10 

SrC 0 3 . 

25 

[Sr+ 2 ]X[C03- 2 ] 

2.0X10 -9 

SrC 2 0 4 . 

18 

[Sr+ 2 ]X[C 2 0 4 - 2 ] 

4.8X10-8 

SrCr 0 4 . 

18 

[Sr +2 X][Cr 0 4 —2 ] 

3.6X10- 5 

SrF 2 . 

i8 

[Sr +2 ]X[F -1 ] 2 

2.8X10- 9 

SrS 0 4 . 

18 

[Sr +2 ]X[S 0 4 ~ 2 ] 

2.7X10- 7 

CaC 0 3 . 

16 

[Ca+ 2 ]X[CO s - 2 ] 

2.8X10- 9 

CaC 2 0 4 . 

18 

[Ca+ 2 1 X[C 2 0 4 - 2 ] 

1.7X10“ 9 

CaCr 0 4 . 

18 

[Ca +2 ]X[Cr 0 4 -2 ] 

0.9X10— 3 

CaF 2 . 

18 

[Ca +2 ]X[F -1 ] 2 

3.4X10- 11 

CaS 0 4 . 

18 

, [Ca +2 ]X[S 0 4 - 2 ] 

6.1X io -5 

MgCOa. 

12 

[Mg+ 2 ]X[CO s - 2 ] 

2.6Xio -8 

MgC 2 0 4 . 

18 

|Mg +2 ]X[C 2 0 r 2 l 

8.6X10- 5 

MgF 2 . 

18 

[Mg+ 2 )X[F->] 2 

7.1X io -9 

MgNH 4 P 0 4 .. . . 

2 5 

[Mg +2 ] X [NH 4 +I ]X [P 0 4 ~ 3 ] 

2.5X10- 13 


1 Magnesium is not precipitated with the other elements of this group by 
the scheme of analysis employed, but on account of the similarity of many of 
its properties, it is considered along with there elements. 


























REACTIONS OF THE BASE-FORMING CONSTITUENTS 91 


Ammonium hydroxide precipitates magnesium hydroxide only. 
The alkaline earth salts differ in general mainly in their 
solubility, schemes of analysis being based on this fact. The 
periodic system and the table of solubility products (p. 90) 
should be studied in order to understand the reactions of 
this group. 

The alkaline earth salts all have colorless bivalent cations. 

REACTIONS OF THE ALKALINE EARTHS 

- 1. Moisten any compounds of barium, strontium, and cal¬ 
cium, separately, with cone. HC 1 , dip into each mixture the 
tip of a clean platinum wire, and hold in the non-luminous 
flame of a burner. Observe the colors imparted to the flame. 

HC1 is added in order to convert any non-volatile compounds into 
the more volatile chlorides. 

Add to separate solutions of salts (nitrates) of barium, 
strontium, calcium, and magnesium, the following reagents. 
Record results with reactions involved. 

2. Ammonium hydroxide. If a precipitate forms, add 
ammonium chloride and explain the result. 

3. Ammonium carbonate. The insoluble carbonates dis¬ 
solve in very weak acids such as acetic acid. 

4. Ammonium oxalate. Test the solubility of the pre¬ 
cipitated oxalates in acetic acid. 

5. Sulphuric acid. 

6. Calcium sulphate, 

7. Potassium chromate. 

8. Ammonium hydroxide and sodium acid phosphate. 

In testing the magnesium salt, add ammonium chloride (to prevent 
the precipitation of magnesium hydroxide). With excess ammonia, 
the neutral phosphates of barium, strontium, and calcium are precipi- 


92 


QUALITATIVE CHEMICAL ANALYSIS 


tated. With magnesium salts, the precipitate consists of magnesium- 
ammonium phosphate, MgNH 4 P 0 4 . 

9. Barium hydroxide. (Test only a magnesium salt with 
this reagent.) 

10. Dissolve some dry CaC03 in HNO3 in an evaporating 
dish, evaporate to dryness and heat until all moisture is 
gone, but do not ignite. Reduce the dried nitrate to a 
powder with a pestle, add a mixture of equal parts of abso¬ 
lute alcohol and ether (this mixture is very inflammable and 
should be kept away from flames), triturate the mixture, 
and filter. Add a little H2SO4 to the filtrate, and observe 
the precipitate of CaSCU (CaSC>4 is quite insoluble in alcohol). 

Repeat the experiment, using BaC03 instead of CaC03. 
Barium nitrate is insoluble in the anhydrous alcohol-ether 
mixture. (This is also true of strontium nitrate.) 

ANALYSIS OF GROUP IV 

66. Boil the filtrate from Group^ III for several minutes 
and filter from sulphur. Add an excess of both NH4OH and 
(NH 4 ) 2 C 0 3 , heat gently for several minutes, and filter. Re¬ 
serve the filtrate for the analysis for magnesium and the 
alkalies. 

Ammonium salts in general slightly increase the solubility of the 
precipitated carbonates, but excess of ammonium carbonate decreases 
the solubility by mass action. 

Excess NH 4 OH decreases the solubility of the carbonates by suppress¬ 
ing the following hydrolysis: 

(NH 4 ) 2 C 0 3 +H 2 0 <=± NH 4 HC 0 3 +NH 4 0 H. 

Ammonium bicarbonate would yield acid carbonate ions HCO _1 3 , which 
would form soluble alkaline earth bicarbonates. 

On long standing in the cold, crystals of magnesium ammonium car- 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 93 

bonate MgC0 3 - (NH^aCOs^^O, may be formed by (NH 4 ) 2 C0 3 , there¬ 
fore the solution should be filtered immediately. 

Ammonium salts prevent the precipitation of magnesium hydroxide. 

67 . Identification of Barium, Strontium, and Calcium. 
Method I. Dissolve the precipitate of BaCOs, SrC03, and 
CaC03, placed in an evaporating dish, in 5 to 10 c.c. HNO 3 . 
Evaporate to dryness and heat with a small flame until all 
HNO3 and moisture are expelled. 

The nitrates must not be heated to such a temperature that they are 
decomposed into oxides. 

Reduce the nitrates to a powder, add 10 c.c. absolute 
alcohol—ether mixture, triturate with a pestle, and filter as 
rapidly as possible through a small filter. 

68 . Add 2-3 drops H2SO4 to the filtrate. A precipitate 
of CaSC>4 is a test for calcium. Filter, wash with alcohol, 
dissolve the residue of CaS04 in a little water, and add 
(NH4)2C204. A white precipitate of CaC204 is a con¬ 
firmatory test for calcium, and distinguishes CaS04 in the 
H2SO4 precipitate from any SrS04 which may be present due 
to imperfect treatment of the nitrates. The CaC204 may be 
dissolved in HC 1 and the red flame test for calcium be 
obtained. 

69 . Wash the residue of Ba(N03)2 and Sr(NOs)2 from 68, 
with alcohol-ether mixture until the filtrate shows no precipi¬ 
tate of CaS04 with a drop of dilute H2SO4; dissolve the 
residue in 50 c.c. of water, and filter if necessary. To a 
small portion of the filtrate add 3 drops of acetic acid, boil, 
and add K 2 Cr04 very slowly; if there is a precipitate, add 
a few drops of acetic acid to the entire filtrate, boil, and add 
K2Cr04 gradually until in excess. A yellow precipitate of 
BaCrC>4 is a test for barium. Filter, dissolve the residue in 


94 QUALITATIVE CHEMICAL ANALYSIS 

HC1, and confirm barium by the green color imparted to the 
flame. 

SrCr0 4 should not be precipitated in the dilute solution and acetic 
acid further lowers its solubility. 

The solution is boiled and K 2 Cr0 4 is added slowly in order that 
BaCr0 4 , as it precipitates out, w r ill have time to deposit on the first 
small crystals precipitated; thereby larger aggregates will be obtained 
than if it were thrown out of solution all at once in the form of fine 
crystals which would pass through the filter. 

70 . To the filtrate from the BaCrO^ or the solution which 
gives no precipitate with K 2 Cr04, add NH 4 OH and 
(NH 4 ) 2CO3, and heat. A precipitate of SrC03 is a test for 
strontium. Filter, dissolve the residue in HC 1 , and confirm 
strontium by the crimson flame test. 

71 . Method II. Dissolve the precipitate of BaCOs, 
SrCC>3, and CaC03, in a little acetic acid, dilute, and add 
K 2 Cr04 to a small portion of the solution. If a precipitate 
forms, heat the entire acetic acid solution (diluted) to boiling, 
and add K 2 CrC>4 gradually until in excess. A yellow pre¬ 
cipitate of BaCrCU is a test for barium. Filter, dissolve the 
residue in HC 1 and confirm barium with the flame test. 

72 . Divide the filtrate from the BaCrC>4, or the solution 
which gave no precipitate with K 2 Cr04, into two parts. To 
one portion add CaSC>4, and boil for several minutes. If a 
precipitate does not form immediately, allow to stand for ten 
minutes. A white precipitate of SrSCU is a test for strontium. 
(Observe the solution with care, as a small white precipitate 
might be obscured by‘the orange-red color of K 2 Cr 2 07). 
Filter, moisten the residue with HC 1 , and confirm strontium 
by the crimson flame test. 

73 . From the second portion of the filtrate from the 
BaCr04 or the solution in which K 2 Cr 0 4 gave no precipitate, 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 95 


remove strontium salts by adding (NH^SCU. Boil, allow 
to stand for ten minutes, and filter from SrSC>4. (A little 
CaS04 may also be precipitated.) To the filtrate add 
(NH 4 )2C204. A white precipitate of CaC204 is a test for 
calcium. Filter, dissolve the residue in HC 1 , and obtain the 
red flame test for calcium. 

While some SrS0 4 may be in the filtrate from the (NH 4 ) 2 S0 4 pre¬ 
cipitate, it is not precipitated by ammonium oxalate (see table of solu¬ 
bilities). 

74 . Identification of Magnesium. Add to a small portion 
of the filtrate from the precipitated carbonates in 66 or to 
the solution in which (NH4)2CC>3 produced no precipitate, 
some ammonium sulphate. A small white precipitate will be 
BaSC>4. To another small portion, add some ammonium 
oxalate, warm, and allow to stand for ten minutes. A white 
precipitate will be CaC204- (The presence of barium and 
strontium in these precipitates may be confirmed by filtering 
and applying the flame test to the residues.) If there are 
traces of barium and calcium in the filtrate from the (NELt^COa 
precipitate as shown by the above tests, add a few drops of either 
(NH4)2SO4 or (NH 4 ) 2 C 2 04 or both to the entire filtrate, allow 
to stand for a few minutes and filter. 

These traces must be removed since barium and calcium would pre¬ 
cipitate in the test for magnesium. If the solution is allowed to stand 
for a considerable length of time and much magnesium is present, am¬ 
monium-magnesium oxalate may crystallize out. 

To the filtrate from 66 free from traces of barium and 
calcium by the above procedure (any trace of strontium is 
allowed to remain) add ammonium-hydrogen phosphate, 
(NH 4 ) 2 HP 0 4 . A white, crystalline precipitate of magnesium- 
ammonium phosphate is a test for magnesium. If the pre- 


96 


QUALITATIVE CHEMICAL ANALYSIS 


cipitate does not immediately form, stir thoroughly, and allow 
to stand in the cold for an hour. 

If the ammonium phosphate precipitate is floccuient, it is probably 
aluminium phosphate, A1P0 4 (carried through by use of too much 
NH 4 OH in precipitating the Group III metals). In such a case, filter, 
and pour acetic acid repeatedly over the residue on the filter, whereby 
MgNH 4 P0 4 is dissolved, but not A1P0 4 . Add NH 4 OH to the acetic 
acid solution. A white crystalline precipitate of MgNH 4 PO* will be a 
test for magnesium. 


Group V. The Alkali Group 
K +1 , Na +1 , NH 4 +1 

75 . The valence of the metals of this group is one. The 
oxides of potassium are gray potassium oxide, K 2 0 , and yel¬ 
low potassium superoxide, K2O4; those of sodium are white 
sodium oxide, Na 2 0 , and yellow sodium peroxide, Na 2 0 *. 
The basicity of the hydroxides decreases in the following 
order, KOH>NaOH>NH 4 OH (cf. p. 165). KOH and 
NaOH are called fixed alkalies, and NH 4 OH volatile alkali 
because of its well-known instability and property of giving 
off ammonia. 

REACTIONS 

i. Heat KC 1 and NaCl on the end of a clean platinum 
wire in the non-luminous flame, and observe the colors. Also 
view the flames through several pieces of cobalt glass. (The 
platinum wire may be cleaned by dipping into HC 1 , holding 
in the flame, and repeating this until no color is imparted to 
the flame.) 

Test KC 1 and NaCl in the Bunsen flame before the spectro¬ 
scope in the dark room. 


REACTIONS OF THE BASE-FORMING CONSTITUENTS 96 a 


Spectrum Analysis. When ordinary white light passes through a 
triangular glass prism the wave lengths of which the light is composed 
are refracted unequally, those of the red the least and of the violet the 
most. The colors emerge from the prism as a series ranging from red, 
orange, yellow, green, blue, to violet, presenting to the observer a continuous 
spectrum. If now the light from a sodium flame, e.g., is passed through 
the prism, only those colors (mainly yellow) of which this light is composed 
will appear. 

A spectroscope is used for spectrum analysis, an ordinary form of this 
instrument consisting essentially of a triangular glass prism and three 
telescopes. One of the latter is the collimator which has a vertical slit 
at one end through which the rays from the substance emitting light are 
passed. These rays pass through a lens at the other end of the collimator 
and emerge as parallel rays which are refracted by the prism. The 
spectrum is seen through an observing telescope placed at the proper 
angle to be in the path of the rays. The image of the slit of the collimator 
is seen clearly and sharply defined as a series of vertical lines. This 
interrupted or line spectrum is characteristic of any element or stable 
compound which is being heated to such a point that it emits light. Light 
from an incandescent lamp or glowing metal as platinum wire will present 
a continuous spectrum since it contains all the wave lengths of white 
light. 

The third tube is a short telescope at the outer end of which is a scale 
(often divided into millimeters) illuminated by a small luminous flame, 
the rays from which pass through a lens at the other end of the telescope 
and emerge as parallel rays which are totally reflected from the prism 
into the observing telescope so that the image of the scale is superimposed 
on that of the line spectrum. The relative positions of the lines obtained 
from any substance may hence be read and compared with those produced 
by known pure substances. 

For the observation of the flame spectrum (the spark spectrum must 
be used for substances emitting light with difficulty in the Bunsen flame) 
dip a platinum wire into a concentrated solution of the substance or suspend 
on the platinum loop some of the dry substance, and heat in the non- 
luminous Bunsen flame. Place the flame 4 to 6 inches from the collimator 
slit and low enough so that the blue cone is out of the line of vision. The 
sodium line will usually appear without the intentional addition of any 
sodium salt, since practically all substances contain sufficient of such 
impurity to give the test. (Sodium may be detected in the spectroscope 
when onl> one part in ten billion is present.) 


966 


QUALITATIVE CHEMICAL ANALYSIS 


Some of the most characteristic and easily observed lines which may 
thus be obtained are presented in the following table: 


Element. 

Line Spectrum. 

Sodium 

Yellow (D) line, (actually consists of two distinct yellow lines) 

Potassium 

Two bright red and two faint violet lines 

Lithium 

Red line 

Calcium 

One bright orange, and one green line, and others less distinct 

Strontium 

Several red lines; one blue line 

Barium 

Many orange and green lines 


In addition to the above methods of spectrum analysis, we have - 
those based on a study of absorption spectra. These methods are partic¬ 
ularly applicable to gases and vapors of heated metals and to colored 
solutions. Very valuable results have thus been obtained, as in the 
knowledge acquired concerning the elements present in the gaseous envelope 
of the sun, the chemistry of dyes, of colored glasses, of the blood, etc. 
In this method the gas or liquid, in a suitable glass container, is placed 
between a source of white light and the collimator slit of the spectroscope. 
There is obtained an absorption spectrum, consisting of dark lines on a 
bright field in place of the bright lines of the emission spectrum, the 
positions of these dark lines being likewise characteristic. In the case of 
colored solutions, instead of the dark lines there are obtained dark ab¬ 
sorption bands whose width is a function of the concentration of the solu¬ 
tion concerned. 

For the purposes of qualitative analysis the spectroscope is of value 
in the detection of minute traces of substances which could not be readily 
identified by other methods. This is particularly true of the alkali and 
alkaline earth group metals. No idea of the quantity of these substances 
is obtained, so that the student should note that spectrum analysis may 
be used as an aid to the longer analytical procedures, but in general 
should not be substituted for them. 

2. Heat separately in a porcelain dish (below a red heat) 
chlorides of potassium, sodium, and ammonium. 

To 5 c.c. each of normal solutions of potassium, sodium, 
and ammonium chlorides add the following reagents: 









REACTIONS OF THE BASE-FORMING CONSTITUENTS 97 


3. A few drops of hydrochlorplatinic acid, H^PtCle. 

The precipitate obtained with the potassium salt, if filtered off and 
gently ignited, will yield a residue which colors the flame violet: 

K 2 PtCl G ->2KCl+Pt+2Cr 2 . 

The precipitate obtained with the ammonium salt, if filtered off and 
ignited, will give off ammonium chloride: 

(NH 4 )zPtCl s - 2 NH 4 CI + 2 CI 2 +Pt. 

4. Add an equal volume of a saturated solution of tartaric 
acid, H2C4H4OG. Shake the mixtures and allow to stand. 
Acid tartrates are precipitated. 

5. A few drops of perchloric acid, HCIO4. 

6 . Add an equal volume of sodium cobaltinitrite solution, 
Na3Co(N02)6, to the solutions of potassium and ammonium 
chlorides. 

The precipitates are K 2 NaCo(N0 2 ) 6 *H 2 0 and (NH 4 ) 2 NaCo(N0 2 )< ; -H 2 0, 
respectively. 

For the above precipitates to form, the solutions must be neutral 
or weakly acid (acetic acid). Why would alkali prevent the action? 

7. Add to the sodium chloride solution an equal volume 
of potassium pyroantimonate solution, K2H2Sb207, scratch 
the sides of the test-tube with a stirring rod and allow to 
stand. 

The precipitate is sodium pyroantimonate. The test must be made 
in neutral or alkaline solution since in acicf solution pyroantimonic acid 
is precipitated. 

8. To the ammonium chloride solution add Nessler’s re¬ 
agent (cf. p. 46). 

Appreciable amounts of ammonia produce a brown precipitate while 
traces are indicated by a yellow color. The reaction may be written: 

2K 2 HgI 4 +3KOH +NH4OH Hg0NH 2 HgI+3H 2 0+7KI. 


98 


QUALITATIVE CHEMICAL ANALYSIS 


ANALYSIS OF GROUP V 

76 . Evaporate the filtrate from Group IV to dryness in 
an evaporating dish, and volatilize the ammonium salts by 
careful ignition of the residue. The desired action is com¬ 
plete when white fumes cease to come off. Add 3 c.c. cone. 
HC1 and repeat to ensure the entire removal of the am¬ 
monium salts. 

Add 5 c.c. cone. HC 1 and evaporate just to dryness (re¬ 
peat this procedure once) so that the metaphosphates of 
sodium and potassium, formed by ignition, may be recon¬ 
verted into the orthophosphates. The metaphosphates are 
difficultly soluble in water. 

Dissolve the residue in the least possible amount of water 
and filter if necessary. Apply the flame test to the solution, 
and if a brilliant, yellow flame, continuous for some time, is 
obtained, the presence of appreciable amounts of sodium is 
indicated. Observe the flame through cobalt glass to obtain 
the violet flame for potassium. 

Divide the solution into two portions (as these will be 
small in amount they may be placed in watch-glasses resting 
on a black surface for the better observation of results). 

77 . To one portion of the solution add a few drops of 
HCIO4. A white precipitate of KCIO4 is a test for potassium. 
Filter and apply the flame test. 

78 . To the other portion of the solution add an equal 
volume of potassium pyroantimonate solution. Stir and allow 
to stand for an hour (or longer) if no precipitate forms imme¬ 
diately. A white crystalline precipitate of Na2H2Sb207 is a 
test for sodium. Filter, dissolve in HC 1 and apply the flame 
test. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 99 

79 . To the original unknown (placed in a small beaker) 
add a little solid calcium oxide if it is in solution, or a little 
calcium oxide and water if the unknown is a solid. Over the 
beaker place a watch-glass on whose lower side is a piece 
of moist red litmus paper, and heat gently. The odor of 
ammonia and the blue color of the litmus paper are tests for 
ammonium. 

REACTIONS OF THE ACID-FORMING CONSTITUENTS. 1 

80 . In the analysis of a substance, the base-forming con¬ 
stituents are determined before the acid-forming constituents. 
In this way many anions are often excluded. Further, pre¬ 
liminary tests indicate the presence of certain anions and spe¬ 
cial tests make the analysis for these much shorter than a 
regular procedure. Many of the acids may be tested for in 
the presence of others, so that a perfectly systematic method 
for the determination of the anions is not always necessary. 
The most practical methods for acid analysis are given in 
Part II, p. 142. 

For convenience in classification the acids are divided into 
the following groups. On account of differences in solubility 
of the salts mentioned, special precautions have to be taken 
in certain cases in the application of these principles (Part 
11, p. 144). 

Group I. The barium salts are insoluble in dilute HNO3. 

Group II. The barium salts are soluble in dilute HN 0 3 , 
but are insoluble in acetic acid. 

Group III. The barium salts are soluble in both nitric 
acid and acetic acid, but are insoluble in neutral solution. 

1 The grouping of the acids is based largely on principles given in Bottger’s 
“ The Principles of Qualitative Analysis.” 


100 


QUALITATIVE CHEMICAL ANALYSIS 


Group IV. The silver salts are insoluble in dilute HNO3. 

Group V. The barium and silver salts are soluble in 
water. (Silver nitrite is only sparingly soluble; it is readily 
soluble in dilute nitric acid.) 

In referring to the acids, it is, perhaps, needless to recall 
to the student that the anhydrous compounds, strictly speak¬ 
ing, should not be called acids; they become acids only when 
dissolved in an ionizing medium. 

Group I. The Sulphuric Acid Group 
S 0 4 - 2 , (SiFo- 2 ) 1 

SULPHURIC ACID, H2SO4 

81 . The valence of sulphur in sulphuric acid is six. 
Hydrogen sulphate is a colorless oily liquid. It begins to 
boil at about 290°, the boiling-point rising to 338°, and it 
decomposes partially into its anhydride in the process. It 
vaporizes much below (about 160°) the temperature at which 
it boils giving off white suffocating fumes. Concentrated sul¬ 
phuric acid is very hygroscopic and takes up water when 
exposed to the air. This power of taking up water will be 
illustrated later in its action on organic matter. Sulphuric 
acid is a dibasic acid yielding the colorless ions HSO4 -1 and 
SO4- 2 . 

The difficultly soluble salts (see table of solubilities) are 
those of calcium, strontium, lead, and barium, besides basic 
sulphates such as those of mercury and bismuth. 

1. Heat any dry sulphate with Na2CC>3 on charcoal. 
Place the fused mass on a silver coin, moisten with water, 
and observe the deposit on the silver. (Hepar reaction.) 

1 Anions of not very common occurrence are placed in parentheses. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 101 

2. Add BaCl 2 to dilute H 2 S 0 4 and then add a little HC 1 . 
BaS 0 4 is somewhat soluble even in dilute HC 1 . 

The fact that BaS 04 is slightly soluble in dil. HC1 may be explained 
by the solubility-product principle, 

Ba +2 XSOr 2 =A, 

the S0 4 2 ion concentration being diminished by the formation of 
HS0 4 -1 ions by the mass effect of the added hydrogen ions: 

H +1 +S0 4 “ 2 ^±HS04" 1 . 

Heat a mixture of BaS 0 4 and cone. H 2 S 0 4 , decant the 
supernatant liquid, and pour this into water. 

The precipitate is BaS0 4 . Explain by the solubility-product prin¬ 
ciple and the fact that the secondary ionization of sulphuric acid is 
small when the acid is concentrated. Assume that Ba(HS0 4 ) 2 is formed 
when the barium sulphate dissolves in the acid. 

hydrofiAjosilicic acid, H 2 SiF 6 

82 . Hydrofluosilicic acid (fluosilicic acid) is formed by the 
action of excess hydrofluoric acid cn silicon tetrafluoride 
(cf. p. 108) in aqueous solution, and is known only in such 
solution: 

H 2 F 2 +SiF 4 ^H*SiF 6 . 

Barium,' sodium, and ammonium fluosilicates are difficultly 
soluble in water. Heated with concentrated sulphuric acid, 
these salts evolve hydrofluoric acid. 


102 


QUALITATIVE CHEMICAL ANALYSIS 


Group II. The Oxalic Acid Group 
Cr 0 4 - 2 , C2O4- 2 , SO3- 2 , S2O3- 2 , F- 1 

CHROMIC ACID, H 2 Cr 04 

(Cf. p. 74) 

83 . 1. Heat some ammonium dichromate in the closed 
tube. 

Dichromates usually give off oxygen and leave a residue of a normal 
chromate with some chromic oxide (K 2 Cr 2 0 7 ), or they may as in this 
case leave only chromic oxide: 

(NH 4 ) 2 Cr 2 07 —» Cr 2 C>3 +N 2 +4H2O. 

2. Add cone. H2SO4 to some K 2 Cr 2 07 and heat. 

The dichromate is reduced to a chromium salt and oxygen is evolved. 

3. Treat a solution of K2Cr0 4 with dil. H2SO4. Then 
make alkaline with NaOH. 

Explain the changes of color and write the reactions. 

4. Dilute some K2Cr0 4 solution, add a little HNO3 and 
hydrogen peroxide, and ether. Shake the mixture and ob¬ 
serve the color of the ether layer. 

This color is due to the presence of the unstable perchromic acid 
H 3 Cr 0 7 . An excess of either the chromate or hydrogen peroxide viti¬ 
ates the test. 

OXALIC ACID, H2C2O4 

84 . Anhydrous oxalic acid is a white amorphous solid 
which crystallizes from water with two molecules of water 
of crystallization, H 2 C204-2H 2 0, this compound being very 
stable at room temperatures. It is a dibasic acid with color- 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 103 

less ions C2O4 2- and HC 2 0 4 _1 , forming neutral and acid salts, 
and also peracid salts such as potassium tetroxalate, 

KHC 2 0 4 H 2 C 204 - 2 H 2 0 . 

It is a weaker acid than hydrochloric acid, but is stronger 
than acetic acid. Since alkaline earth salts are more or less 
insoluble in ammoniacal solution, the presence of the oxalate 
radical in a substance necessitates the taking of certain pre¬ 
cautions in the analysis for the Group III bases for the same 
reasons as the presence of the phosphate radical. 

Oxalates, except those of magnesium and the alkalies, are 
insoluble in water. Many insoluble oxalates form soluble 
complex salts in the presence of an excess of alkali oxalate. 
Thus ammonium ferrioxalate, (NH 4 )3Fe(C 2 0 4 )3, is soluble 
in water and gives no tests for the ferric ion. This, again, 
must be borne in mind in the analysis of the Group III 
bases. 

All oxalates are decomposed by ignition, carbon monoxide 
being given off. For example: 

CaC 2 0 4 —> CaC 0 3 +C 0 ; 

on continued heating, the carbonate decomposes: 

CaC 0 3 —> CaO-f C 0 2 . 

1. Heat a dry oxalate with cone. H 2 S 0 4 . Test the com¬ 
bustibility of the escaping gas. Hold a stirring rod, on the 
end of which is a drop of Ba(OH) 2 solution, over the mouth 
of the test-tube. 

2. Add CaCl 2 and BaCl 2 , separately, to solutions of am¬ 
monium oxalate, and then acidify with acetic acid. 

From the table of solubilities, which reagent would give a better 
test for the oxalate ion? 


104 


QUALITATIVE CHEMICAL ANALYSIS 


3. To a solution of ammonium oxalate add AgNOs- 
Test the solubility of Ag2C204 in HC2H3O2 and in HNO3. 

4. To a solution of ammonium oxalate acidified with dilute 
sulphuric acid add a little potassium permanganate and warm 
the mixture. Note the change of color and the evolution of 
a gas: 

2KMn04+5H2C2O4+3H2SO4 —* K2S04 + 2MnS04 

-I-8H2O + 10CO2. 

Work out this reaction on the dualistic basis. 

This reaction is not a specific test for the oxalate ion, as sulphurous 
acid, hydrosulphuric acid, etc., also reduce permanganates. 

Oxalic acid may be detected in the presence of other members 
of Group II in the following way : Add hydrochloric acid and alcohol to 
the solution to be tested, and boil. Thereby chromates are reduced 
to chromium salts with the evolution of aldehyde, C 2 H 4 0 2 : 

K 2 Cr 2 0 7 +3C 2 HX>+8HCl -> 2KCl+2CrCl 3 +3C 2 H 4 0 2 +7H 2 0; 

sulphites are decomposed with the evolution of sulphur dioxide; thio¬ 
sulphates are decomposed with the evolution of sulphur dioxide and the 
deposition of sulphur. Boil until the alcohol is all volatilized and filter 
from sulphur. Make the filtrate alkaline with ammonium hydroxide 
and filter from the precipitated chromium hydroxide. Acidify with 
hydrochloric acid and add barium chloride. Filter from BaS 0 4 . Make 
the filtrate alkaline with ammonium hydroxide, acidify with acetic acid, 
and add a little calcium chloride (since calcium oxalate is more insoluble 
than the barium salt). The precipitate may be an oxalate or a fluoride 
or both (although the fluorides are somewhat soluble in excess of am¬ 
monium salts). Filter, wash the residue, and dissolve in dilute sul¬ 
phuric acid. Test this solution for oxalic acid with potassium perman¬ 
ganate as in Exp. 3. 


SULPHUROUS ACID, H2SO3 

85 . Sulphurous acid is known only in solution since such 
solutions readily decompose into water and the anhydride 
sulphur dioxide: 


H2SO3 H2O+SO2. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 105 


It is a dibasic acid with the colorless ions HSC^ -1 and SO3 -2 , 
forming neutral and acid salts which are unstable, especially 
in solution, slowly oxidizing in the air to sulphates; alkali 
sulphites are therefore best made for immediate use by 
saturating water with sulphur dioxide and neutralizing the 
solution with alkali hydroxide. 

The alkali sulphites are soluble in water; normal sulphites 
of other bases are insoluble, but in some cases, as with cal¬ 
cium, the sulphites dissolve in an excess of sulphurous acid, 
forming acid sulphites. 

Sulphites (as do most sulphur compounds) after fusion 
with Na2CC>3 on charcoal, blacken silver (Hepar reaction). 

Use a neutral solution of an alkali sulphite for the follow¬ 
ing tests. 

1. Heat with dilute H^SCU and note the odor of the gas 
evolved. 

Sulphites are decomposed by nearly all the common acids except 
boric and carbonic acids. 

2. Treat with BaCl 2 and a little acetic acid. The pre¬ 
cipitate is BaS03. Distinguish the precipitate from BaS04 
by adding HC 1 and observing the result. - 

3. Treat with AgNCV Test the solubility of the Ag 2 SC>3 
in HC2H3O2 and in HNO3. 

4. Add a little mercurous nitrate solution. 

Mercury is precipitated and the sulphite is oxidized to a sulphate. 

Sulphurous acid is quite a strong reducing agent. It decolorizes 
iodine solutions (reduces the iodine to hydriodic acid) and potassium 
permanganate solutions. It reduces chromates to chromium salts, and 
iodates to iodine (with excess further to hydriodic acid). 


106 


QUALITATIVE CHEMICAL ANALYSIS 


THIOSULPHURIC ACID, H2S2O3 

86 . Thiosulphuric acid, with the colorless ions S2O3 -2 , is 
known only in solution and is very unstable, breaking up 
.into sulphurous acid and sulphur: 

H 2 S 203 -*H 2 S 03 +S. 

Solutions of sodium thiosulphate, Na2S203, become tur¬ 
bid after standing some time exposed to the air, due to 
liberation of thiosulphuric acid by carbonic acid and decom¬ 
position of the former. 

Nearly all thiosulphates are soluble in water (the barium, 
silver, and lead salts are only sparingly soluble). 

Use a solution of sodium thiosulphate for the following 
tests. 

1. Warm with cone. H2SO4. 

2. Add a little BaCl2 and then a little acetic acid. 

The precipitate is BaS 2 0 3 . 

3. To AgNC>3 add thiosulphate in excess. 

The silver salt first precipitated dissolves in excess, forming a com¬ 
plex salt: 

Ag2S203"l-2Na2S203 —» Na4Ag2(S203) 3 . 

The precipitated silver thiosulphate, if allowed to stand, will darken, 
due to decomposition: 

Ag 2 S 2 03+H20 —> Ag 2 S+H2SO4. 

4. Add a little iodine solution to the thiosulphate. 

The latter is oxidized to sodium tetrathionate: 


2Na 2 S 2 0 3 +l2—* 2NaI+Na2S 4 OG. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 107 


Chlorine and bromine react with thiosulphate to form sulphuric 
acid: 

Na 2 S20,-fH 2 0-fBr2 —► 2NaBr+H 2 S0 4 +S. 

Sulphates, sulphites, and thiosulphates may be detected in the 
presence of each other as follows: Acidify the solution with HC1 and 
add BaCl 2 in excess. Filter off the BaS0 4 (test for sulphate). To the 
filtrate add an excess of iodine solution. Any sulphite is oxidized to 
sulphate: 

H2SO3+H.O+I2 -> 2 ffl+H 2 S 0 4 , 

and thiosulphate is oxidized to a tetrathionate as above. A precipitate 
of BaS0 4 is therefore a test for the presence of a sulphite since barium 
tetrathionate, BaS 4 O t „ is soluble. Filter, and to the filtrate add bromine. 
The tetrathionate is converted into sulphate and a precipitate of BaS0 4 
at this point is a test for the presence of a thiosulphate. 


HYDROFLUORIC ACID, H 2 F 2 

87 . Hydrogen fluoride is a colorless liquid boiling at 19.4°. 
At room temperatures its vapor density indicates that the 
molecular formula is H 2 F 2 . Dissolved in water it acts as a 
weak acid, yielding the colorless fluoride ions F _1 . So- 
called acid salts, such as KHF 2 , are very common. 

Alkali fluorides are soluble in water, the solutions being 
slightly alkaline (hydrolysis); the alkaline earth fluorides are 
insoluble; those of copper, lead, zinc, and ferric iron are 
slightly soluble; fluorides of silver (note difference from other 
halogen salts of silver) and mercury are freely soluble. 

1. Mix some dry calcium fluoride with cone, sulphuric 
acid n a platinum crucible or lead dish. Over the con¬ 
tainer place a watch-glass whose under side is coated with 
paraffin on which several marks have been drawn to expose 
portions of the glass. Warm the mixture gently, allow to 


108 


QUALITATIVE CHEMICAL ANALYSIS 


stand for several minutes, remove the paraffin from the 
watch-glass, and note the etching. 

CaF 2 +H 2 S 0 4 -> CaS 0 4 +H 2 F 2 , 

CaSiOs (one of the constituents of glass) +3H 2 F 2 —» SiF 4 

+CaF 2 +3H 2 0. 

2. Warm a mixture of calcium fluoride, silica (sand), and 
cone, sulphuric acid in a small test-tube. Hold over the 
mouth of the test-tube a drop of water in the loop of a 
platinum wire and note the development of a turbidity (silicic 
acid). 

The H 2 F 2 formed by the action of sulphuric acid on the fluoride 
(cf. Exp. i) acts on silica thus: 

Si0 2 + 4 HF —► SiF 4 -{-2H 2 0, 

and the volatile silicon tetrafluoride reacts with water: 

3 SiF 4 + 4 H 2 0 -> Si(OH) 4 + 2 H 2 SiF 6 . 

3. Add BaCl 2 to a solution of potassium fluoride (or any 
soluble fluoride) and acidify with acetic acid. 

The precipitate is BaF 2 . 

Group III. The Phosphoric Acid Group 
P 0 4 - 3 , AsOs" 3 , As0 4 " 3 , B 0 2 _1 , Si 0 3 “ 2 , CO3- 2 , C 4 H 4 0g~ 2 

PHOSPHORIC ACID, H 3 P 0 4 

88 . Phosphorus, with a valence of three or five in its 
compounds, forms the oxides P 2 0 3 and P 2 05, which are the 
anhydrides of the phosphorous and phosphoric acids respect¬ 
ively. Orthophosphoric acid, H 3 P 0 4 , pyrophosphoric acid 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 109^ 

H4P2O7, metaphosphoric acid, HPO3, hypophosphoric acid, 
H4P2O4, phosphorous acid, H3PO3, and hypophosphorous 
acid, H3PO2, all form salts. Only orthophosphoric acid or 
ordinary phosphoric acid will be considered here. 

Anhydrous phosphoric acid is a translucent, deliquescent 
solid of M P. 41.7 0 . When heated it is converted first into 
pyrophosphoric acid, and then into metaphosphoric acid. 
It forms three classes of salts, primary, secondary, and ter¬ 
tiary. Primary sodium phosphate, NaH2P04, in solution is 
weakly acid in reaction, showing that the ion H 2 P04 _1 is 
further dissociated: 

H2PO4- 1 <=± h +i +hpo 4 - 2 . 

Secondary sodium phosphate reacts alkaline in solution on 
account of hydrolysis: 

Na 2 HP 0 4 +H 2 0 «=± Na 0 H+NaH 2 P 0 4 . 

Tertiary sodium phosphate, Na3P04, is stable only when dry, 
as it is strongly hydrolyzed. 

Alkali phosphates and primary alkaline earth phosphates 
are soluble in water. Other phosphates are insoluble. 

1. Treat a solution of Na 2 HP04 with BaCl 2 , and then add 
an excess of acetic acid. 

The precipitate is BaHPG 4 , soluble only in a large excess of acetic 
acid. 

BaCl 2 precipitates Ba 3 (P0 4 )2 from soluble phosphates in the presence 
of ammonium hydroxide. 

Explain how these facts indicate that the secondary and tertiary 
dissociation of phosphoric acid is very low. 

2. To Na 2 HP04 add a little FeCU. 

The precipitate is FeP0 4 , insoluble in acetic acid. 


110 


QUALITATIVE CHEMICAL ANALYSIS 


Is FeP 0 4 more or less soluble in water than Ba 3 (PC) 4 ) 2 ? What 
has this fact to do with the separation of the Group III metals in the 
presence of phosphoric acid? 

3. To Na2HP04 add AgNOs. Test the solubility of the 
precipitated Ag 3 P04 in acetic acid and in nitric acid. 

Phosphoric acid is identified in systematic analysis in the filtrate 
from the Group II metals by the formation of yellow ammonium phos- 
phomolybdate with ammonium molybdate in nitric acid solution. In 
this way it is tested for in the absence of arsenic; this is desirable, 
since arsenates give a similar precipitate of yellow ammonium arseno- 
^molybdate with ammonium molybdate (longer heating of the solution 
is required for precipitation). 

The phosphate ion may also be identified by reaction with magnesia 
mixture in ammoniacal solution, a white crystalline precipitate of mag¬ 
nesium ammonium phosphate being obtained. This precipitate may be 
distinguished from the similar magnesium ammonium arsenate by 
dissolving in HC 1 and passing in H 2 S. Only the arsenate, of course, 
would give a precipitate of As 2 S 6 , 


ARSENIOUS ACID, H 3 As 0 3 , AND ARSENIC ACID, H3ASO4 
(Cf. p. 54) 

89 . 1. To solutions of potassium arsenite and arsenate 
add BaCl 2 , and test the solubility of the precipitated salts 
in acetic acid. 

2. Repeat Exp. i, using AgN 0 3 instead of BaCl 2 . Test 
the solubility of the precipitates in nitric acid. 


Arsenites may be distinguished from arsenates by their reactions 
with silver nitrate, with magnesia mixture, with ammonium molybdate, 
or with H 2 S in acid solution. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 111 


BORIC ACID, H3BO3 

90 . Boron has a valence usually of three in its compounds. 
Its oxide, B2O3, is the anhydride of orthoboric acid, H3BO3, 
a colorless, crystalline solid, not very soluble in water and 
showing only weakly acid properties. It is slightly ampho¬ 
teric, acting as a base towards certain acids; thus BPO4 is 
known. When heated, orthoboric acid loses water and yields 
metaboric acid, HBO2, and pyroboric acid, H2B4O7. 

Salts of metaboric acid and pyroboric acid are known, 
pure salts of orthoboric acid being unknown. 

Borax, Na2B 4 07* 10H2O, may be used in the same way 
as microcosmic salt as a bead test for certain metals. Heated, 
it loses water, and the pyroborate combines with basic oxides 
to form borates. 

Borax, the most common salt containing boron, reacts 
alkaline in solution: 

Na2B407+3H 2 0 2NaB02 + 2H3B03. 

If the solution is very dilute, complete hydrolysis takes 
place: 

NaB02 + 2H20 NaOH+HsBOs- 

1. To a little dry borax in a porcelain dish add a few 
drops of cone, sulphuric acid and some alcohol. Set fire to 
the alcohol and observe the color shown at the edge of the 
flame. 

This color is that of burning ethyl borate, (CiHOsBOs. 

Na 2 B 4 0 7 +H2SO4 + 5H2O —> Na 2 S 0 4 +4H3BO3, 

3C 2 H 6 OH +H 3 BO 3 -* (C 2 H5) 3 B03 -f 3H2O. 

Halides of copper should be absent, as they give the same test. 


112 


QUALITATIVE CHEMICAL ANALYSIS 


2. Add BaCl2 to a concentrated solution of borax, and 
test the solubility of the precipitated barium metaborate, 
Ba(B02)2, in acetic acid. 

The presence of boric acid would cause the same difficulty in the 
separation of the Group III metals from the alkaline earths as do 
phosphoric acid and hydrofluoric acid; but the borates of the alkaline 
earths are not very insoluble in water, and are especially soluble in the 
presence of ammonium salts, so that boric acid rarely causes any com¬ 
plications in the regular procedure. 

3. Add AgN03 to a concentrated solution of borax. Test 
the solubility of the precipitated AgBC>2 in acetic acid. 

4. Dip a piece of turmeric paper in a solution of borax 
made weakly acid with HC 1 , and observe the color of the 
paper when it has dried. 

silicic acid, LLSi0 4 

91 . Silicon has a valence of four in its compounds, and 
its oxide, Si02, is the anhydride of the silicic acids, of great 
number and corresponding to various stages of hydration. 
The amphoteric nature of H 4 Si 0 4 (assumed to be the formula 
of ordinary silicic acid, although the composition of the 
silicic acids and of silicate ions is very indefinite) is indicated 
by the fact that' SiCl 4 and similar compounds are known, 
although they hydrolyze in water: 

SiCl 4 + 4 H 2 0 —MHCl+Si(OH) 4 . 

The student is referred to other text-books for facts con¬ 
cerning the solubility of various silicates. A water-soluble 
silicate will yield a precipitate with BaCl 2 or AgNC>3 and these 
precipitates are transformed by acids, silicic acid being formed. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 113 

It remains in solution or is precipitated and may set to a 
jelly. 

1. Test a grain of sand in the microcosmic salt bead. 

The so-called “ skeleton ” of undissolved silica floating in the melted 
bead is ordinarily a characteristic test for silica and silicates. 

2. Test the reaction of a solution of water-glass, Na2Si03, 
to litmus. ' 

3. Add HC 1 gradually to a solution of sodium silicate 
until the solution reacts acid. 

After some time, or immediately, gelatinous silicic acid may separate: 
Na 2 Si 0 3 + 2HCI +HoO 2NaCl+Si(OH) 4 . 

4. Pour a little sodium silicate solution into some cone. 
HC 1 . The silicic acid remains in the colloidal condition. 
Make the solution alkaline with NH4OH. 

5. Fuse some sand (Si02) with three or four times its 
volume of Na 2 CC>3 in a platinum dish. 

Na 2 C 0 3 +Si 0 2 -> Na 2 Si 0 3 +C 0 2 . 

Boil the fusion mixture with water and acidify with HC 1 . 
Silicic acid may or may not precipitate at this point. Evap¬ 
orate to dryness in either case and heat the residue for thirty 
minutes at 125 0 . The silicic acid is thereby converted into 
a less hydrated form which is insoluble in HC 1 . Add HC 1 
and note the insolubility of the residue. 

Recall Exp. 2 under “ Hydrofluoric Acid,” p. 108. This test serves 
as a test for silica as well as for hydrofluoric acid. 


114 


QUALITATIVE CHEMICAL ANALYSIS 


CARBONIC ACID, H2CO3 

92 . Carbon usually has a valence of four in its compounds. 
Carbon dioxide is the anhydride of the hypothetical carbonic 
acid, H2CO3, known only in solution. Carbonic acid is a 
very weak dibasic acid, and its alkali salts react alkaline in 
solution because of hydrolysis. It yields the colorless ions 
HCO3- 1 and CO3- 2 . 

The normal carbonates of the alkalies are alone soluble 
in water. Acid carbonates of other metals, especially of the 
alkaline earths, are also soluble. 

1. To Na 2 C03 in a test-tube add a little HC 1 , and hold 
the end of a stirring rod on which is a drop of barium hydrox¬ 
ide solution over the mouth of the test-tube. 

In testing small quantities of carbonates, a blank experiment should 
be tried, since the laboratory air may contain considerable amounts of 
carbon dioxide from illuminating gas flames. 

TARTARIC ACID, H2C4H4O6 

93 . Tartaric acid is a colorless, crystalline solid which 
decomposes on ignition. The neutral alkali tartrates and 
sodium acid tartrate are soluble in water. Ammonium and 
potassium acid tartrates and other tartrates are difficultly 
soluble in water. It yields the colorless ions HC 4 H 406 -1 
and C4H4O6 -2 . 

The interference of tartaric acid and similar compounds 
in the analysis of the metals has been discussed. 

1. Heat some Rochelle salt (sodium-potassium tartrate) 
in the closed tube and observe all phenomena—odor, evolu¬ 
tion of carbon dioxide (detected by Ba(OH)2), deposit of 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 115 


moisture on the sides of the tube, and blackening due to car¬ 
bon. Treat the cold residue with HC 1 and note the effer¬ 
vescence. N 

Organic substances are compounds of carbon and generally behave 
in the same way as Rochelle salt. However, a residue of metallic car¬ 
bonate from which carbon dioxide can be liberated by HC 1 as above, 
is only obtained if there is an alkali metal in the compound. The 
charring and the odor of burnt sugar is characteristic of organic com¬ 
pounds. Carbonates and oxalates do not show this effect, and are 
usually considered along with inorganic compounds. 

2. Warm a mixture of Rochelle salt and cone, sulphuric 
acid. 

Note the carbonization, and evolution of C 0 2 and of CO, which burns 
with a blue flame, the burnt sugar odor, and the odor of sulphur 
dioxide (evolved by the reduction of the sulphuric acid). 

3. To a solution of Rochelle salt add AgN03. The pre¬ 
cipitated Ag2C4H406 is soluble in acids and ammonia. Add 
dilute ammonium hydroxide, drop by drop, until the precipi¬ 
tate dissolves. Place the test-tube in water heated to 70° 
and allow to stand for fifteen minutes or until a metallic 
mirror of silver appears, due to the decomposition of silver 
tartrate. 

4. BaCb precipitates barium tartrate from solutions of 
Rochelle salt, and the precipitate is soluble in acetic acid. 

Ammonium salts also render the alkaline earth tartrates more 

soluble. 

To a solution of Rochelle salt add CaCl2 in slight excess. 
(Calcium tartrate is soluble in acetic acid, thereby distin¬ 
guishing it from calcium oxalate.) Filter and treat the resi¬ 
due of calcium tartrate in the cold with a strong solution of 


116 . QUALITATIVE CHEMICAL ANALYSIS 

NaOH free from carbonates. Dilute the solution containing 
the dissolved tartrate (it is present probably as the complex 
salt, Na2C 4 H 2 Ca04) and filter (from CaC 0 3 ) if necessary. 
Boil the solution and note the reprecipitation of calcium 
tartrate. 

5. Prepare a saturated solution of tartaric acid and 
neutralize it with NaOH. Acidify the solution with acetic 
acid, add some potassium acetate and stir and allow to 
stand in the cold if a precipitate of potassium acid tartrate, 
KHC 4 H 4 Og, does not form immediately. 

Carbonates may be distinguished from oxalates and tartrates by 
the evolution of carbon dioxide with dilute acids. To distinguish 
oxalates from tartrates the property which tartrates possess of forming 
complex anions with copper is utilized (cf. p. 71). To the solution 
made alkaline with NaOH add a little copper sulphate and filter from 
any undissolved copper hydroxide. In the presence of the tartrate ion 
the filtrate will show a deep blue color due to the presence of the com¬ 
plex copper-tartrate anion. This test must be carried out in the absence 
of ammonium salts (which in NaOH solution would form the blue cop¬ 
per-ammonium ions) and of arsenites (which precipitate copper arsenite, 
soluble in excess alkali). Arsenic may be removed by precipitation with 
hydrogen sulphide. Ammonium salts may be decomposed with the 
evolution of ammonia by boiling the NaOH solution before the addi¬ 
tion of copper sulphate. 

To detect carbonates in the presence of sulphites, treat the 
unknown with acetic acid. Carbonates are quickly decomposed, while 
sulphites decompose only slowly. Pass the gas evolved through a 
delivery tube into lime water. The precipitate formed may be filtered 
and again treated with acetic acid. Note the odor of the gas evolved, 
by which CaC 0 3 may be distinguished from CaS 0 3 . 

Carbonates may also be distinguished from sulphites by boiling the 
unknown solid or solution with a concentrated solution of potassium 
dichromate, and passing the evolved gas into lime water. K 2 Cr 2 0 7 
oxidizes any sulphite to sulphate. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 117 


Group IV. The Hydrochloric Acid Group 

Cl- 1 , Br- 1 , I- 1 , CN->, S- 2 FeCCN),;- 4 , Fe(CN) e - 3 , SCN- 1 , 
CIO3- 1 , (BrOs- 1 , IO3- 1 ) 

HYDROCHLORIC ACID, HC 1 ; HYDROBROMIC ACID, HBr; HYDRI- 
ODIC ACID, HI 

94 . Many of the reactions of these acids may be under¬ 
stood if it is recalled ( 28 ) that the electro-affinity of the halo¬ 
gens varies in such a way that the strength of their ions 
decreases according to the series F>Cl>Br>I. The free 
halogen will assume the charge of the weaker ion— chlorine, 
e.g., liberates bromine from bromides. 

The chlorides, bromides, and iodides of mercurous mercury, 
silver, cuprous copper, and lead, and mercuric iodide and 
certain basic salts of the halogens are insoluble in water. 
(Consult table of solubilities.) 

1. Warm some dry alkali chloride, bromide, and iodide, 
separately, with cone, sulphuric acid. 

Observe all the results and write all reactions. This experiment 
illustrates the slight reducing power of hydrobromic acid and the strong 
reducing power of hydriodic acid. 

2. To solutions of alkali chloride, bromide, and iodide, add 
separately AgN 0 3 . Acidify with HN 0 3 and then add an 
excess of NH 4 OH. 

Note the color of the precipitated silver salts and the difference in 
their solubility in ammonium hydroxide. 

To detect chlorides in the presence of bromides and iodides, 
to 1 c.c. of the HN 0 3 solution add a few drops of AgN 0 3 (side-shelf 
reagent, N/5, diluted with twice its volume of water). Boil, allow to 
settle or filter if necessary, and to the clear solution add a few drops 


118 


QUALITATIVE CHEMICAL ANALYSIS 


more AgN0 3 . Continue in this manner until all the yellow Agl and the 
yellowish-white AgBr are precipitated, after which white AgCl is pre¬ 
cipitated if chloride is present. 

3. The following experiment illustrates a method for de¬ 
tecting a chloride in the presence of bromide and iodide. 

Mix equal portions of solutions of alkali chloride, bromide 
and iodide; add lead dioxide and acetic acid, and boil until 
all free bromine and iodine are expelled. 

At this point there is no odor of the free halogens, and if the mix¬ 
ture is allowed to settle, the supernatant liquid is colorless. 

Halogen acids yield a plumbic salt and free halogen with lead dioxide. . 

Pb0 2 + 4 HC1 —>PbCl 2 +Cl 2 4-2H,0. 

Filter, and wash the residue with hot water to dissolve 
lead chloride. To the filtrate add HNO3 and AgNC>3. White 
AgCl indicates the presence of chloride. 

If the precipitate is yellow, the lead dioxide mixture was not boiled 
sufficiently. 

Another method for the detection of chlorides in the presence 
of bromides and iodides is as follows: To the dry substance mixed with 
dry potassium dichromate, add cone. H 2 S0 4 , and lead the vapors given 
off in the reaction, into NH 4 OH. To the ammoniacal solution add 
acetic acid and lead acetate. Yellow PbCr0 4 is precipitated in the 
presence of chlorides. 

Chromium oxychloride is given off as a reddish gas: 

4 NaCl+K 2 Cr 2 0 7 +3H 2 S0 4 - 2 Cr0 2 Cl 2 + 2Na 2 S0 4 +K 2 S0 4 -f3H 2 0, 
and colors the ammonium hydroxide yellow: 

Cr0 2 Cl 2 + 4 NH 4 0H -> (NH 4 ) 2 Cr0 4 + 2 NH 4 C1 + 2 H 2 0. 

Bromine and iodine are liberated from their salts similarly, but yield 
colorless salts with NH 4 OH. 

4. To detect bromides and iodides in the presence of each 
other (and of chlorides) mix solutions of the three alkali 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 119 


halides. To about 3 c.c. of the mixture add a few c.c. 
carbon disulphide and then chlorine water very gradually. 
A violet coloration will first appear, showing the presence of 
the iodide. On further addition of chlorine the iodine is 
oxidized to colorless iodic acid, HI0 3 , and the carbon disul¬ 
phide becomes red or brownish-red, due to the liberation of 
bromine. Any further excess of chlorine will form a colorless 
compound with the bromine. 

5 . To a little alkali iodide solution add H2SO4 (dilute 
sulphuric acid must be used so that the action may not be 
attributed to the sulphuric acid alone) and a little sodium 
nitrite, NaN 02 (solid). Observe the color of the solution. 

Traces of iodide may be detected by this reaction by adding starch 
to the reaction mixture. 

2HI + 2HNO2 -> 2N0+I 2 + 2 H 2 0 . 

. Chlorides and bromides are not oxidized by nitrous acid in the cold. 

Note the ease with which the iodide ion loses its charge and forms 
iodine. 

HYDROCYANIC ACID, HCN 

95. Hydrogen cyanide is a colorless liquid of boiling-point 
26 °. It dissolves in water, yielding the very weak acid hydro¬ 
cyanic acid, with colorless CN -1 ions. 

Alkali, alkaline earth, and mercuric cyanides are soluble 
in water, while other cyanides are insoluble. The formation 
of complex soluble cyanides with certain metals has been 
indicated previously. 

1 . To some silver nitrate solution add potassium cyanide, 
little by little until present in excess. 

Recall reactions of silver salts (p. 39). 


120 QUALITATIVE CHEMICAL ANALYSIS 

2 . Add a few c.c. ferrous sulphate solution and ferric 
chloride solution to a few drops of potassium cyanide solu¬ 
tion. Make the mixture alkaline with NaOH, heat for a 
few minutes and acidify with HC1. 

In the presence of traces of cyanide, a green solution instead of a 
blue precipitate is obtained. To explain this test, recall the reactions 
of salts of iron (p. 69). 

HYDROSULPHURIC ACID, H2S 

96. Hydrogen sulphide, a colorless poisonous gas, dissolves 
in water to form the weak dibasic acid hydrosulphuric acid, 
H 2 S, with the colorless ions SH _1 and S -2 . The aqueous solu¬ 
tion slowly becomes turbid on standing in the air: 

H 2 S+0-*H 2 0+S. 

Sulphides of the alkalies and of strontium and barium are 
soluble in water. Other sulphides are insoluble. While cab 
cium sulphide is insoluble in water as such, it decomposes, 
forming the soluble acid salt, calcium hydrosulphide: 

2 CaS + 2H 2 0 -> Ca(OH) 2 + Ca(SH) 2 . 

Dilute and concentrated sulphuric acid decomposes almost 
all sulphides and H 2 S is evolved. This may be best detected 
by holding over the test-tube a strip of filter paper moist¬ 
ened with an alkaline solut on of a lead salt (in NaOH, 
sodium plumbite is therefore the reagent). The formation 
of black PbS is a very sensitive test. Certain sulphides 
(arsenic sulphide, mercuric sulphide, etc.) are best decom¬ 
posed by boiling with cone. HC1 and a little metallic zinc. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 121 


HYDROFERROCYANIC ACID, H 4 Fe(CN)e; HYDROFERRICYANIC 

acid, H 3 Fe(CN 6 ). 

97. Both these compounds :’n the anhydrous condition are 
solids which dissolve in water to form a tetrabasic and a 
tribasic acid respectively. 

The alkali and alkaline earth ferrocyanides are soluble in 
water (barium only slightly); other ferrocyanides are insolu¬ 
ble (except certain mixed salts). 

The alkali and alkaline earth ferricyanides are soluble in 
water. Almost all other ferricyanides are insoluble. 

The behavior of salts of these acids on ignition may be 
illustrated by the following reactions: 

K 4 Fe(CN ) 6 -> 4 KCN+FeC 2 +N 2 , 

2 K 3 Fe(CN ) 6 -> 6 KCN+ 2 FeC 2 + 2 N 2 + (CN) 2 . 

They react with cone. H 2 S0 4 as follows: 

K 4 Fe(CN) 6 +6H 2 S0 4 +6H 2 0 -> 2 K 2 S0 4 +FeS0 4 

+3 (NH 4 ) 2 S 0 4 +6CO, 

2 K 3 Fe(CN) 6 + 1 2 H 2 S0 4 + i 2 H 2 0 -> 3 K 2 S0 4 +Fe 2 (S0 4 ) 3 

+ 6 (NH 4 ) 2 S 0 4 +i 2 C 0 . 

Study reactions of iron (p. 70 ) to recall the formation of 
Prussian blue and Turnbull’s Blue. 

1 . To solutions of potassium ferrocyanide and potassium 
ferricyanide add AgN0 3 and then acidify with HN0 3 . 

The precipitates are Ag 4 Fe(CN)e and Ag 3 Fe(CN)j, respectively. 

SULPHOCYANIC ACID, HSCN 

98. This is an unstable liquid, dissolving in water to form 
a weakly acid solution with the colorless ions SCN -1 . 


122 


QUALITATIVE CHEMICAL ANALYSIS 


Silver, mercury and copper sulphocyanates are insoluble 
in water. Almost all other sulphocyanates are insoluble. 

Recall the reactions of sulphocyanates with ferric salts 
(p- 7®) • 

1. Add AgNOs and HNO3 to a solution of potassium 
sulphocyanate. 

The precipitate is AgSCN. 

For the properties and reactions of hypochlorite, iodate, and bromate 
ions, the student is referred to larger text-books. 


Group V. The Nitric Acid Group 

N03- 1 , NO2- 1 , C2H3O2- 1 , CIO3- 1 , Mn 0 4 ~ 1 

NITRIC ACID, HNO3 

99. Anhydrous nitric acid ( 99.8 per cent) is a colorless 
liquid which begins to boil at 86 ° with decomposition: 

4HNO3 —» 4NO2 + 2H2O+O2. 

It dissolves in water to form the strong acid, nitric acid, which 
has the colorless ons N 03 -1 . The oxidizing power of nitric acid 
has been illustrated many times in the previous experiments. 

All its salts are soluble in water except a few which are 
partially hydrolyzed to basic salts. (Basic nitrates of bis¬ 
muth and mercury are insoluble.) 

1 . Heat some potassium nitrate in a closed tube. 

KN 0 2 is left as a residue. Ignition to a high temperature would 
convert potassium nitrite into the oxide. 

2 . Heat a little potassium nitrate with cone. H2SO4. 

2HNO3 H2O+N2O4+O. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 123 


3. Dilute a few drops of potassium nitrate solution to 
2 c.c., add 2 c.c. FeS04, cool the mixture thoroughly, incline 
the test-tube at an angle, and pour carefully down the side 
some cone. H2SO4 in such a way that it forms a layer below 
the solution of the nitrate. 

A brown ring caused by the formation of a very unstable com¬ 
pound FeS0 4 -N0, should be observed. The nitric oxide, NO, results 
from the reduction of the nitric acid by the ferrous sulphate. The NO 
then combines with excess of ferrous sulphate. Write all the reactions 
involved. 

This test would be obscured by the presence of a bromide, iodide, 
or sulphocyanate (why?), so that to detect nitric acid in the presence 
of these ions, the solution is shaken with a little silver sulphate, Ag 2 S0 4 ; 
after filtering, the filtrate may be tested for HN0 3 as above. 

4. Boil some sodium nitrate solution with sodium hydrox¬ 
ide and a little zinc dust. Hold a moist piece of red litmus 
paper above the test-tube. 

Ammonia and sodium zincate are formed by this reaction. 

Nitric acid may be detected in the presence of chloric acid by this test. 

4. Treat 1 c.c. potassium nitrate with 4 c.c. cone. H2SO4 
and add 2 c.c. brucine solution. (The brucine solution is a 
0.2 per cent solution of brucine in cone. H2SO4.) The red color 
which appears at first will gradually fade to a light yellow. 

This test allows the detection of nitric acid in the presence of nitrous 
acid. It must be borne in mind, however, that very slight traces of 
nitric acid give the test. 


NITROUS ACID, HNO2 

100 . Nitrous acid is known only in solution and is very 
unstable: 

3HNO2 — > HNO3 + 2NO+H2O. 

All nitrites with the exception of silver nitrite, potassium 


124 QUALITATIVE CHEMICAL ANALYSIS 

cobaltic nitrite, and certain basic nitrites, are soluble in 
water. 

1. Add cone. H2SO4 to some sodium nitrite. 

2. Add AgN03 to some sodium nitrite solution and then 
acidify with HN 0 3 . The precipitate is AgN 0 2 . 

Nitrous acid may be detected in the presence of nitric acid by the 
oxidation of hydriodic acid (p. 119 ). 

ACETIC ACID, HC2H3O2. 

101 . Anhydrous acetic acid (glacial acetic acid) is a color¬ 
less, crystalline solid, with a melting-point of 16.5°. Dis¬ 
solved in water it yields a weak monobasic acid with the 
colorless ions C2H3O2 -1 . 

Silver and mercurous acetates are slightly soluble in water. 
Certain basic acetates are insoluble, while other acetates are 
soluble. 

1. Heat some dry sodium acetate in a closed tube. 

2. Warm some sodium acetate with cone. H2SO4 and note 
the odor of acetic acid. 

3. Add ferric chloride solution to a solution of sodium 
acetate. 

The color is due to the presence of the slightly dissociated ferric 
acetate. The presence of a sulphocyanate would obscure this test 
(why?). If the solution is diluted and boiled, the presence of an ace¬ 
tate would be detected by the precipitation of basic ferric acetate. 

4. To a mixture of equal volumes of cone. H2SO4 and 
ethyl alcohol add some solid sodium acetate and heat gently. 

The odor is due to the formation of volatile ethyl acetate: 


C 2 H 6 0H+HC 2 H 3 0 2 <=* C 2 H 5 C 2 H 3 0 2 +H 2 0. 


REACTIONS OF THE ACID-FORMING CONSTITUENTS 125 

5 . Heat a little solid sodium acetate with a minute quan¬ 
tity of arsenic trioxide in the closed tube and observe the 
odor, which is that of cacodyl oxide: 

4NaC2Hs02 + AS2O3 —> [As(CH 3 ) 2 ]20+2Na 2 C0 3 + 2CO2. 

Do not inhale much of the fumes as cacodyl oxide is intensely poison¬ 
ous. (Other organic acids related to acetic acid also give this test.) 


CHLORIC ACID, HCIO3. 

102 . Chloric acid is known only in a hydrated form, since 
attempts to isolate it result in decomposition: 

3HCIO3 -> HCIO4 + 2C 10 2 +H 2 0 . 

Salts of chloric acid, with the colorless ions CIO3 -1 are 
all soluble in water. 

1. Heat some solid KCIO3 in the closed tube. Test the 
gas evolved with a glowing splinter. Dissolve the residue 
in the tube in water and add AgN03 and HNO3. 

2 Heat some KCIO3 on charcoal. 

The combustion of the charcoal due to the presence of the strong 
oxidizing agent is called deflagration. This effect should be distin¬ 
guished from decrepitation such as is observed on igniting crystals of 
NaCl, which may be disintegrated, due to the volatilization of mechan¬ 
ically enclosed moisture. 

3. Warm a small amount of solid KCIO3 with cone. H2SO4. 

The explosive gas evolved is chlorine dioxide, C10 2 . 

4. To some KCIO3 add dilute H2SO4 and ferrous sulphate 
solution, and boil. Add some potassium ferrocyanide. 

Since the ferrous salt has been oxidized, how would the presence of 
chlorates interfere with the “ring” test for nitrates because of the 
above reaction? 


126 QUALITATIVE CHEMICAL ANALYSIS 

To detect chlorates in the presence of nitrates, the chlorate may be 
reduced to chloride with FeS0 4 as above, and the chloride tested for 
with AgNC> 3 . 

PERMANGANIC ACID, HM11O4 

Permanganic acid is known only in aqueous solutions. Its 
salts, with the reddish-violet permanganate ions MnCU -1 , are 
all soluble in water. Since permanganates are readily re¬ 
duced to manganous salts by H 2 S in acid solution, manganese 
is detected in basic, analysis and the color indicates the 
presence of permanganate. 


PART II 


SYSTEMATIC ANALYSIS OF UNKNOWN SUBSTANCES 
PRELIMINARY TESTS 

103. In the systematic analysis of a given unknown, it is 
advantageous to carry out certain preliminary tests, whereby 
many helpful suggestions for methods of dissolving the sub¬ 
stance, if it is a solid, are obtained, and indications of com¬ 
binations of the constituents and particularly of the presence 
of organic matter are revealed. 

104. If the substance is a solid, a qualitative examina¬ 
tion of its physical properties—color, odor, specific gravity— 
should be made. (Preliminary tests are not necessary in the 
analysis of metals and alloys.) If the unknown is a liquid, 
note its color, odor, and reaction to litmus. An acid reaction 
indicates the presence of free acid, an acid salt, or a hydro¬ 
lyzed salt; an alkaline reaction indicates a free base or a 
hydrolyzed salt. Evaporate some of the.liquid to dryness, 
and if there is a residue, subject portions of it to the follow¬ 
ing tests. Treat solid unknowns directly according to these 
tests, first reducing them to a fine powder if necessary. 

105. Heat a little of the substance in the closed glass tube. 
The indications furnished by this test are given on page 128 . 

Especial attention must be paid in the test to any indi¬ 
cations of the presence of organic matter, since it must be 
removed, if present, before analysis of the substance in the 
wet way is undertaken. 


127 


128 


QUALITATIVE CHEMICAL ANALYSIS 


Observation 


Indication 


(A) A sublimate forms. 

Transparent.. 


White, easily volatile. 

White, difficultly volatile. 

Yellow or reddish brown. 

Yellow. 

Yellow, red when rubbed. 

Gray, forming globules when rubbed.. 

Black, accompanied by a garlic odor.. 

Black, red when rubbed./ 

Black, accompanied by a violet vapor. 
(B) A gas is evolved. 

0 *, inflaming a glowing splinter. 

Odor of NH 3 . 


Odor 9 f H 2 S and blackening of lead acetate 

paper. 

Odor of S0 2 . 

CO, burning with a blue flame. 

CO 2 , rendering Ba(OH) 2 turbid. 

The halogens, recognized by their odor and 

color. 

Odor and reddish brown color of N0 2 . 

(C) The substance undergoes a change of 
color and the residue becomes: 

Black, accompanied by a burnt odor. 

Black, no burnt odor. 

Brownish red while hot, yellow when cold.... 

Reddish brown. 

Red or black while hot, red when cold. 

Green. 

Orange while hot, yellow when cold. 

Yellow while hot, white when cold. 

Yellow. . 


Water of constitution or crystalli¬ 
zation, or hygroscopically or 
mechanically held. 

Ammonium salts, As20 3 , Hg 2 Cl 2 , 

HgCl 2 . 

Sb 2 0 3 . 

S'. 

As 2 S 3 . 

Hgl 2 . 

Hg, from the metal or its com¬ 
pounds. 

As. 

HgS. 

I. 

Chlorate, alkali nitrate, peroxide, 
etc. 

Ammonium salts, cyanides, or¬ 
ganic matter containing nitro¬ 
gen. 

Hydrated sulphides, etc. 

Sulphites, thiosulphates, etc. 

Oxalates, etc. 

Carbonates, oxalates, and organic 
matter. 

Chlorides, bromides, and iodides. 

Nitrates and nitrites. 


Organic matter. 

Copper, cobalt, nickel, etc. 
Lead. 

Cadmium. 

Iron. 

Chromates. 

Bismuth. 

Zinc. 

Tin. 


































PRELIMINARY TESTS 


129 


106 . Heat some of the substance on charcoal before the 
oxidizing flame of the blowpipe. 


Observation 

Indication 

(A) The substance deflagrates. 

(B) The substance fuses and is absorbed 

by the charcoal. 

(C) A white infusible residue is obtained. 

Moisten and test reaction to litmus.. . . 

Chlorates, nitrates, peroxides. 

Salts of the alkali metals. 

An alkaline reaction indicates the 
alkaline earths. A neutral re¬ 

Moisten with Co(N0 3 )2 and ignite. 

action indicates Al, Mg, 01 Zn. 

With Co(N 0 3 ) 2 , yellowish green 
indicates Zn, blue indicates Al, 
silica, or phosphates, pink in¬ 
dicates Mg. 

(D) An incrustation is formed: 

Yellow when hot or cold. 

Yellow while hot, white when cold. 

Reddish brown. 

White, very volatile, odor of garlic. 

White, less volatile. 

(E) Odor of SO 2 . 

Pb, Zn. 

Zn, Sn. 

Cd. 

As. 

Sb. 

S. 

- T- 


107 . Heat some of the substance with about twice its 
weight of sodium carbonate on charcoal before the reducing 
flame of the blowpipe. Incrustations will be observed as in 
106 . Test the malleability of any metallic bead formed. 
In order to detect certain small metallic particles, grind the 
heated mass with water in a mortar, and decant off the lighter 
particles of carbon and soluble matter. 


Observation 

Indication 

(A) A white, metallic, malleable bead is 

(B) A gray, brittle, metallic bead is formed. 

(C) Red metallic particles or a red metallic 

bead results. 

(D) An infusible, black, magnetic powder 

Pb, Ag, Sn. 

Sb, Bi. 

Cu. 

Fe, Co, Ni. 



























130 


QUALITATIVE CHEMICAL ANALYSIS 


108. Heat a small particle of the substance in the micro- 


cosmic salt bead in the Bunsen flame. Heat first in the 
oxidizing flame and then in the reducing flame. Continue 
heating until all action has ceased. 

Observation 

Indi cation 

Undissolved particles swimming in the bead 
(“ skeleton ”). 

Silica. 

In oxidizing flame: 

Green when hot; blue, cold. 

Copper. 

Cobalt. 

Blue. 

Reddish brown while hot; colorless or light 
yellow, cold. 

Nickel, Iron. 

Amethyst. 

Manganese. 

Chromium. 

Green. 

In reducing flame: 

Red. 

Copper. 

Cobalt. 

Blue. 

Brownish or yellow. 

Nickel. 

Reddish while hot; greenish, cold. 

Iron. 

Green. 

Chromium. 



109. Moisten some of the substance with a little concen¬ 
trated hydrochloric acid and try the flame test, using platinum 
wire. If sodium is present, view the flame through cobalt 
glass. 


Observation 

Indication 

Brilliant yellow color lasting some time. 

Violet. 

Sodium. 

Potassium. 

Calcium. 

Strontium. 

Lithium. 

’Arsenic, antimony, lead. 

Barium. 

Zinc. 

Boric acid, manganese. 

Copper. 

Brick red. 

Crimson. 

Carmine. 

Blue to violet. 

Yellowish green. 

Whitish green. 

Bluish green. 

Emerald green. 




































PRELIMINARY TESTS 


131 


Small quantities of elements may be detected by the use 
of the spectroscope. Such tests, however, give no idea of the 
quantity present. 

110 . Heat a small portion of the substance with a little 
dilute sulphuric acid. Only a small portion of the substance 
should be used on account of the possible presence of cyanides. 


Observation 

Indication 

C0 2 is evolved; gives test with Ba(OH) 2 . 

HCN is evolved with odor of bitter almonds. .. 
SO2 is evolved with characteristic odor; renders 
Ba(OH) 2 turbid. 

Carbonates. 

Cyanides. 

Sulphites. 

Thiosulphates. 

Nitrites. 

Sulphides. 

Alkali and alkaline earth perox¬ 
ides. 

Acetates. 

SO2 is evolved and S is formed. 

Brown N0 2 is evolved with characteristic odor. 
H 2 S is evolved with characteristic odor; gives 
test with PbAc 2 paper. 

0 2 is evolved, inflaming a glowing splinter.... 

HC2H3G2 is evolved with characteristic odor. . . 


111 . To a small portion of the unknown solid substance 
add a little concentrated sulphuric acid, and warm gently if 
necessary (do not heat to such a temperature that the sul¬ 
phuric acid is decomposed). The indications furnished by this 
test are given on page 132 . 

ANALYSIS OF A SOLUTION 

112 . If a solution contains organic matter, evaporate just 
to dryness an amount which is estimated (as is possible 
from the closed tube test) would contain one gram of inor¬ 
ganic matter. Remove the organic matter by the procedure 
given in 113 and proceed with the analysis as there indicated. 

In the absence of organic matter, if the solution is neutral 
or acid in reaction, proceed with the analysis for the base¬ 
forming constituents (Tables I-V). 












132 


QUALITATIVE CHEMICAL ANALYSIS 


Observation 

Indication 


Carbonates, oxalates, tartrates, 


cyanates, and other organic 
matter. 

Oxalates, tartrates, cyanides, and 
other organic matter. 

Tartrates and other organic mat¬ 


HC2H3O2 is evolved, with characteristic odor. . . 

hcn . 

ter. 

Acetates (some charring may 
occur). 

Cyanides. 


Sulphites. 

SO9 and S . 

Thiosulphates. 

N0 2 . 

Nitrates, nitrites. 

h 2 s . 

Sulphides. 


Permanganates, chromates, per¬ 

H 2 F 2 is evolved, with penetrating odor: etches 
glass... . 

oxides. 

Fluorides, fluosilicates. 

HC1 is evolved with sharp odor; reddens moist 
litmus .. 

Chlorides. 

HBr, Br with its characteristic odor and red 

hrnwn rolor . 

Bromides. 

HI, I with its characteristic odor and color; 
SOo and S are also nroduced. 

Iodides. 

l± 20 j u\y 2 ) CHl vl O ciiV/ cviov wwv* ********* 

CI 2 is evolved, with a greenish yellow color, and 

rViarartf>ri<;tir ndnr . 

Hypochlorites. 

Chlorates. 

C10 2 is evolved, with yellow color and explosive 
properties . 


If the solution is alkaline in reaction add an excess of HC1. 
If no precipitate forms, proceed with the analysis for the 
metals. If a precipitate forms, filter, dissolve the residue 
according to methods outlined in 114 for the solution of 
solid substances, and analyze for the metals. 

A precipitate may be an arsenic group metal sulphide held in solu¬ 
tion as an alkali sulpho-salt, silver cyanide held in solution by excess 
potassium cyanide, etc. 

























ANALYSIS OF A SOLID SUBSTANCE 


133 


To analyze for the acid-forming constituents proceed ac¬ 
cording to the methods presented in 117 for a substance solu¬ 
ble in water. 

ANALYSIS OF A SOLID SUBSTANCE FOR THE BASE-FORMING 
CONSTITUENTS 

113 . Organic Matter is Present. If the preliminary tests 
in the closed tube or with cone. H2SO4 have indicated the 
presence of organic matter, place an amount of substance 
estimated to contain one gram of inorganic matter in a por¬ 
celain dish; add about 8 c.c. of a mixture of equal parts of 
cone. H2SO4 and cone. HNO3, and heat gently until thick 
white fumes of su'phuric acid begin to be evolved. If free 
carbon is still present, add more cone. HNO3, heat again 
until H2SO4 volatilizes, and continue this process until the 
free carbon has been completely oxidized. The mixture will 
then no longer be colored black. 

It is unnecessary to carry out the above procedure when carbonates 
or acetates are present, but such organic matter as tartrates, citrates, 
etc., must be destroyed, as it interferes with the regular procedure for 
the analysis of the base-forming constituents, particularly in the pre¬ 
cipitation of the third group trivalent metals. Other kinds of organic 
matter such as starch, gums, etc., interfere with processes of filtration 
and washing. Oxalates are also decomposed, thereby preventing the 
precipitation of the alkaline earth metals along with the third group 
metals. 

Evaporate the decarbonized mixture, as obtained above, 
almost to dryness and dilute with 50 c.c. water. If there is 
no residue proceed with the analysis for the base-forming 
constituents. A residue may often consist of silica or the 
sulphates of lead, barium, strontium, or calcium (if this 


134 


QUALITATIVE CHEMICAL ANALYSIS 


metal is present in excess ). 1 If there is a residue, filter, wash, 
and analyze the filtrate for the metallic ions. Treat the residue 
according to 116 below. 

114. Solution of Non-Metallic Solids (Soluble in Acids) in 
the Absence of Organic Matter. Test the solubility of a very 
small portion of the substance in (a) water, ( b ) dilute HC1, 
(c) cone. HC1, (<0 dilute HN0 3 , (e) cone. HN0 3 , (/) aqua 
regia. If in doubt as to whether any of the substance dis¬ 
solves in a certain solvent, filter, and evaporate the filtrate 
to dryness on a watch-glass; a residue will indicate, by its 
amount, the extent of solution. 

(a) If the substance is soluble in water, dissolve one gram 
in 50-100 c.c. water (if more water is required, the substance 
is considered to be insoluble in this solvent) and analyze for 
the metals according to 112. 

(b) , ( c ) Dissolve one gram of the substance in 10 c.c. dilute 
HC1 or 5 c.c. cone. HC1 and dilute to 50-100 c.c. 

If basic salts such as BiOCl or SbOCl are precipitated on dilution, 
a few c.c. HC1 in excess will redissolve them. If chlorides soluble in 
cone. HC1 such as PbCl. (white needles) or AgCl (white and curdy) 
are precipitated on dilution, the use of HN0 3 as a solvent will probably 
be more practicable. 

If the substance is only partially soluble in water while the 
remainder, after filtering, dissolves in acid, mix small portions 
of the aqueous and acid solutions. If there is no precipitate, 
mix the entire solutions and analyze for the metals. If 
there is a precipitate, analyze the solutions separately for the 
metals. 

1 In this as well as in other observations, every conceivable possibility 
of the presence of combinations of compounds is not called to the mind of 
the student, and it is left to him to develop the most practicable procedure 
for particular cases. 


ANALYSIS OF A SOLID SUBSTANCE 


135 


An example of the latter case would be a mixture of sodium sulphate 
and barium carbonate. Water would dissolve the sodium sulphate, 
and HC1 would dissolve the barium carbonate. If the solutions were 
mixed, barium sulphate would be precipitated. 

Gelatinous silicic acid may separate when the substance is treated 
with acid. In this, or any case when the presence of silicon has been 
indicated by the preliminary tests, some other solvent such as HN0 3 
must be used, since in the removal of silica by dehydration volatile 
chlorides of mercury and arsenic would be lost if HC1 were employed 
as the solvent. 

(i d)j ( e ) If dilute HNO3 or cone. HNO3 is the best solvent 
for the substance, dissolve one gram in 10 c.c. of the former 
or 5 c.c. of the latter. In case a larger amount of dilute 
HNO3 is required or if cone. HN 0 3 is the solvent, evaporate 
the solution just to dryness, add 5-10 c.c. dilute HN 0 3 , 
dilute to 50-100 c.c. and analyze for the metals. 

Excess HNO3 interferes with the precipitation of the sulphides of 
the second group metals because of the oxidation of H 2 S. 

If silicic acid is present, evaporate the HNO3 solution just 
to dryness, and heat on the water-bath for a quarter to half 
an hour to render the silica insoluble. Treat the residue 
with 5-10 c.c. dilute HN 0 3 , dilute to 50-100 c.c., filter, and 
analyze the filtrate for the metals. 

By removing silica at this point, later tests for aluminium are not 
rendered uncertain by the presence of silicic acid. 

HN0 3 oxidizes certain elements from a lower to a higher state of 
oxidation, such as mercury, arsenic, and iron. Cone. HN0 3 may also 
cause sulphur to be precipitated and oxidizes sulphur compounds to 
sulphuric acid, whereby certain insoluble sulphates may be precipitated. 

(/) If aqua regia is used as a solvent, the solution of one 
gram of the substance is evaporated just to dryness to expel 
excess of acid. (If silicic acid is present, dehydrate as in 
(, d ), {e) above, add 5-10 c.c. dilute HC 1 to the residue, 


136 


QUALITATIVE CHEMICAL ANALYSIS 


dilute to 50-100 c.c., filter, and analyze the filtrate for the 
metals.) Add 5-10 c.c. dilute HC 1 , dilute to 50-100 c.c. 
and analyze for the metals. 

If aqua regia dissolves the substance only partially, the 
residue is washed thoroughly and treated as in 116 below. 

115 . Solution of a Metallic Substance. Warm one gram 
of the metal or alloy with dilute HNO3 until action ceases. 
Certain salts may crystallize out at this point and should 
be dissolved in water. 

If the substance is entirely soluble in HNO3, tin is absent 
and only a trace of antimony may be present. Evaporate 
the solution to dryness and heat at 125 0 for a quarter to a 
half an hour. Add 5-10 c.c. dilute HNO3 and dilute to 
50-100 c.c. Filter from the residue, which is silica, and ana¬ 
lyze the filtrate for the metals. 

Of the acid-forming constituents, carbon, sulphur, phosphorus, and 
silicon are most commonly present in metals or alloys. 

A black residue insoluble in HN0 3 indicates the presence of carbon, 
silicon, carbides or silicides of the metals, or noble metals such as gold 
or platinum. 

If there is a white residue insoluble in dilute HNO3, it 
may contain hydrated oxide of antimony, metastannic acid, 
stannic arsenate, stannic phosphate, or silica, Evaporate the 
HNO3 solution just to dryness, add 5-10 c.c. HNO3, dilute 
to 50-100 c.c., filter, and wash the residue. Add HC 1 to the 
filtrate and if a precipitate of PbCk or AgCl forms, filter, 
and analyze the residue according to the regular procedure 
for the metals of the first group. Add the filtrate from the 
precipitated chlorides, or the solution in which HC 1 pro¬ 
duced no precipitate, to the solution of the residue from the 
HNO3 treatment, which solution is obtained as below. 


ANALYSIS OF A SOLID SUBSTANCE 


137 


Dissolve the residue, insoluble in dilute HNO3, in aqua 
regia , evaporate to dryness, add 5 c.c. dilute HC1, and dilute 
to 50 c.c. If there is a residue (of silica) filter. To this 
solution add the HC1 solution obtained as in the previous 
paragraph, and analyze for the metals of Groups II—III. 
As some silicic acid may remain in solution the precipitate ob¬ 
tained in the test for aluminium must be identified according 
to the procedure given on page 86. 

116 . Solution of a Non-Metallic Solid (Organic Matter Being 
Absent) Insoluble in Acids. Any residue undissolved by acids 
( 114 ) or by cone. H2SO4 ( 113 ) may ordinarily contain car¬ 
bon, sulphur, lead chloride, lead sulphate, silver chloride, cal¬ 
cium sulphate, strontium sulphate, barium sulphate, calcium 
fluoride, silica, silicates, fluosilicates, native or ignited oxides 
of iron, chromium, aluminium, or tin, chrome iron ore, 
Prussian blue and related substances. PbBr2, PbD, AgBr, 
Agl, and AgCN would be converted into the corresponding 
chlorides by the action of aqua regia; sulphur may be present 
from the oxidation of some sulphur compound by- the same 
reagent or HNO3 alone; sulphates may have been produced 
by a similar oxidation. 

{a) Carbon and Sulphur. Heat the residue in a porce¬ 
lain crucible. Free carbon will be oxidized to carbon dioxide 
unless it is in the form of graphite. The presence of graphite 
is indicated by the black, greasy character of the residue. 
Sulphur will burn with a pale blue flame and give off the 
characteristic odor of sulphur dioxide. 

(b) Test for Lead Chloride . If the presence of lead or 
silver has been indicated by the preliminary tests, boil the 
residue with water and filter while hot. Test the filtrate for 
lead by adding K 2 CrC>4. 


138 


QUALITATIVE CHEMICAL ANALYSIS 


If lead chloride is present, extract the residue until this 
salt is completely removed. 

(c) Test for Lead Sulphate. Boil the residue with a con¬ 
centrated solution of ammonium acetate slightly acid with 
acetic acid, filter, and test the filtrate for the presence of 
lead by adding K2Cr04. 

If lead sulphate is present, extract it completely from the 
residue. 

(d) Test for Silver Chloride. Heat the residue from which 
lead has been removed (if that metal is present) with a solu¬ 
tion of potassium cyanide, filter, and add NH4OH to the 
filtrate. Pass in H 2 S and if a black precipitate of Ag 2 S 
forms, filter, wash the residue thoroughly, and dissolve it in 
hot dilute HNO3. Dilute the solution and add HC1. 

If silver chloride is present, extract it completely from the 
residue. 

(e) Fusion with Alkali Carbonate in a Platinum Crucible. 
If the presence of lead or silver has not been indicated by the 
preliminary tests, ( b ), (c) and (< d ) may be omitted. Since 
reducible metals readily form alloys with platinum, and 
therefore destroy platinum crucibles, if they are present in 
the substance only the residue obtained after removing them 
should be subjected to the following procedure: 

Fuse such a residue, or one gram of the original substance 
containing no reducible metal, sulphides, or any organic 
matter with four to ten times its weight of a mixture of 
equal parts of sodium carbonate and potassium carbonate, in a 
platinum crucible. (Consult an instructor.) The residue for 
the fusion should be previously dried. A mixture of car¬ 
bonates is used, since it has a lower melting-point than either 
carbonate alone. Heat over the blast lamp for several min- 


ANALYSIS OF A SOLID SUBSTANCE 


139 


utes and if a clear melt is obtained the action is completed. 
If there are undissolved particles left after ten to fifteen min¬ 
utes’ heating, add potassium nitrate, little by little, avoiding 
a large excess. (Fused alkali nitrates slowly attack plati¬ 
num.) 

By fusion with alkali carbonates, sulphates such as barium sulphate 
are converted into carbonates soluble in acid. Silica and all silicates 
are converted into soluble sodium silicate. The amphoteric oxides such 
as aluminium oxide will dissolve more or less in excess of the carbonate. 
By the addition of KN0 3 chromium oxide free or combined as in chrome 
iron ore 'is oxidized to a soluble chromate. Ignited ferric oxide, or the 
native or ignited oxides of aluminium and tin are little affected by the 
fluxes used. 

If the cooled crucible with the fusion is placed in water 
and heated, the fusion may generally be removed from the 
crucible. Grind the fusion in a mortar, boil with water until 
the action has ceased, filter, and wash the residue thoroughly. 
Treat the residue with HN0 3 , heating until the action is 
completed, and filter from any undecomposed residue. Make 
the aqueous extract of the fusion acid with HNO3, mix a 
small portion of this solution with a small portion of the HNO3 
solution of the fusion and if there is a precipitate produced, 
treat the two solutions separately as below for the metals. 

A precipitate would form, for instance, if BaS0 4 had been fused with 
Na 2 C0 3 ; Ba(N0 3 ) 2 , resulting from the action of HN0 3 on BaC0 3 , 
would precipitate BaSCh from the aqueous solution of Na 2 S0 4 . 

If no precipitate is formed and if silicic acid is present, 
evaporate to dryness and heat at 125 0 for twenty minutes. 
Treat the residue with 5 c.c. HNO3, dilute to 50 c.c., and 
filter if there is a residue of silica. Confirm the presence of 
silicic acid by heating the residue with a little cone. H2SO4 
and H 2 F 2 on platinum foil; there should be no residue 


140 


QUALITATIVE CHEMICAL ANALYSIS 


after evaporation. Analyze the solution containing no silicic 
acid or freed from it, for the metals of Groups II-IV. 

Note that tests for the alkalies are excluded or account of the fusion 
with alkali carbonates, and that lead may be present because of its 
existence in the original substance as a silicate. 

(/) Test for Alkalies in Silicates Undecomposed by Acids. 
To test for alkalies in silicates which are undecomposed by 
acids, treat with 5 c.c. cone. H2SO4 in a platinum dish, heat 
on a water bath in the hood, and add 5 c.c. H2F2 little by 
little. (The vapors of H2F2 must not be breathed nor should 
any of it be allowed to come in contact with the skin.) 
Heat until the action ceases, and then heat over the free 
flame until H2SO4 begins to volatilize. Cool, dilute to 50 c.c., 
filter from insoluble sulphates and undecomposed compounds, 
and test for the alkalies in the filtrate after removing metals 
of the first four groups by the various group reagents. 

If there is reason to suppose that a substance, or any part 
of it unattacked by acids, is nothing but silica, SiC>2, treat with 
cone. H2SO4 and H 2 F 2 as above, and evaporate to dryness. 
With silica alone there should be no residue. 

(g) Fusion with Alkali Carbonate in a Porcelain Crucible. 
If it is desired to fuse that part of a substance undecomposed 
by acids, with alkali carbonates and nitrate in a porcelain 
crucible, instead of following the procedure in ( 5 )-(e), this 
may be done, and lead chloride or sulphate, or silver chloride, 
does not need to be previously removed. In such a case, 
if there is reason to suspect their presence, analyze the solu¬ 
tion of the fusion for silver as well as the other metals. Since 
fused alkali carbonates attack porcelain, tEsts for silicic acid, 
aluminium, and calcium in such a solution are futile. 

(Ji) Solution of Fluorides . Fluorides in silicates may be 


ANALYSIS OF A SOLID SUBSTANCE 


141 


decomposed by fusion with alkali carbonate, but fluorides 
such as CaF 2 are not appreciably attacked. In the latter 
case, the residue of the substance undissolved by acids should 
be treated with a little cone. H 2 S 0 4 in a platinum vessel, and 
the mixture evaporated almost to dryness. It is then diluted 
with water and filtered. The filtrate is analyzed for the metals 
(a large amount of CaSOi from CaF 2 would be in solution) 
and any residue is treated as in (b)—(e). 

(i) Solution of Oxides. To dissolve the calcined or native 
oxides of iron and aluminium, which are little affected by acids 
or by fusion with an alkaline flux, they may be heated in a 
platinum crucible with potassium acid sulphate and the 
mixture fused until solution is complete (four to ten times 
the weight of the substance would ordinarily be a sufficient 
amount of the sulphate to use). The oxides are converted 
into sulphates soluble in water or a little HC 1 . Silica and 
silicates are not attacked by fused potassium acid sulphate. 

Stannic oxide may be decomposed by fusing it in a por¬ 
celain crucible with four to ten times its weight of a mixture 
of equal parts of sodium carbonate and sulphur (fused sulphur 
and sulphides attack platinum). The cooled fusion is boiled 
with water, filtered from excess sulphur, and analyzed for 
tin as in the ammonium polysulphide filtrate from the Group 
II sulphides (the stannic oxide has been converted by the 
fusion into a soluble sodium sulphostannate). Antimony tri- 
oxide could be rendered soluble by the same treatment. 

(/) Complex Cyanides. -Such compounds as Prussian blue 
are best decomposed by heating with a solution of sodium 
hydroxide. The cation of the cyanide forms an insoluble 
hydroxide which'may be filtered from the solution of the 
sodium salt containing the anion of the cyanide. 


142 


QUALITATIVE CHEMICAL ANALYSIS 


ANALYSIS OF A SOLID SUBSTANCE FOR THE ACID¬ 
FORMING CONSTITUENTS 

117 . Preparation of Solution for the Analysis for the Acid Ions. 

It is not necessary in all cases that metallic ions should be 
removed from a solution which is to be tested for the acid 
ions, but the following procedure providing for their removal 
is presented as a generally applicable one. 

(a) If the substance contains only metals of the alkaline 
earth and alkali groups and dissolves in water with a neutral 
reaction (affects neither blue nor red litmus), dissolve one gram 
in 50 c.c. water and use this solution for the tests for the 
acids. If it dissolves with an acid reaction, make the solu¬ 
tion of one gram in 50 c.c. water exactly neutral with NH4OH 
(adding the reagent from a dropper and testing after the 
addition of each drop with litmus). Such a solution is ready 
for the analysis for the acids. If the substance dissolves with 
an alkaline reaction, neutralize a solution of one gram of the 
substance in 50 c.c. water with HNO3, filter if necessary, and 
test the filtrate for all acids except HNO3. Treat any pre¬ 
cipitate as in (c) below. Test the original substance for 
HNO3. 

(1 b ) If the substance is soluble in water and contains 
metals of the first three groups, or is insoluble in water, 
but soluble in dilute acids, boil a gram of the substance with 
not too large an excess of sodium carbonate and filter. All 
metals except arsenic and the alkalies are (in general) pre¬ 
cipitated as carbonates, basic carbonates, or hydroxides, and 
the filtrate will contain the sodium salts of the acids. Certain 
phosphates are little affected by the treatment, but H3PO4 
has been tested for in the analysis for the metals. 


ANALYSIS OF A SOLID SUBSTANCE 


143 


If there is no interference from the presence of certain 
other acids, test the orginal substance for HNO3, and neutral¬ 
ize the entire Na 2 CC>3 filtrate as below. If a solution of so¬ 
dium salts is desired to test for HN 0 3 , divide the Na 2 C 0 3 
filtrate into two unequal portions. Acidify the smaller por¬ 
tion with H 2 S 0 4 and test for HN 0 3 . Add to the other por¬ 
tion HNO3, and if a precipitate is formed when the solution 
is nearly neutral, filter. This precipitate may be hydroxides 
of zinc or aluminium, or carbonates such as that of copper, 
held in solution by excess Na 2 C03. Make the solution 
slightly acid with HN 0 3 and boil out C0 2 completely (as it 
would interfere with later tests with BaCl 2 and AgN 0 3 ). 
Make the solution exactly neutral with NH 4 OH and test 
for all anions except nitrate ion, as prescribed in 118 . 

In certain cases, boiling with Na 2 C 0 3 does not remove 
the metals (the presence of tartaric acid prevents the pre¬ 
cipitation of certain Group II metals). In such a case, also 
in general if the substance contains only metals of Group II 
(the following treatment removes these more completely than 
the Na 2 C03 treatment), dissolve the substance in water or a 
slight excess of dilute nitric acid, and saturate the solution 
with H 2 S. Filter, and boil the filtrate until all H 2 S is expelled, 
make the solution neutral with NH 4 OH, and test .'or the acids 
as outlined in 118 . 

On account of the reducing power of H 2 S, in the presence of certain 
oxidizing agents such as nitrates and chlorates it may be useless to test 
for H 2 S0 4 , HC1, and HC10 3 . 

(c) If the substance or part of it is insoluble in dilute 
acids, fuse a gram of it with Na 2 C03 in a porcelain crucible. 
(This may be unnecessary if it has been proved that the 
residue consists of nothing but Si 0 2 , Al 2 03, Sn 0 2 , etc.). 


144 


QUALITATIVE CHEMICAL ANALYSIS 


Boil the fusion with water and filter. Add HNO3 to the 
filtrate, boil out CO2, neutralize the solution with NH4OH, 
and test for the acids as in 118 . 

118 . Analysis for the Acid Ions. The student should note 
that H3PO4 and many amphoteric acids (H3ASO4, HMn04, 
H 2 Cr04, etc.) have .made known their presence in the analysis 
for the metals. Further, H2CO3, H 2 SiC>3, H 2 S, and certain 
other acids, have revealed their presence in the operations 
involving solution of the substance in acids. If the presence 
of Ii 2 CC>3 or H 2 S is doubtful, make special tests for these 
acids with the original substance. The original substance 
should be tested for boric acid, since its barium salt is not 
very insoluble and its presence might not be indicated by the 
group tests below. 

Tests for individual acids are made according to the 
methods described on pages 100-126. 

(a) To a portion of the prepared neutral solution of so¬ 
dium salts, containing the anions as prepared in 117 , add a 
little BaCl 2 and CaCl 2 . 

CaCl 2 is added since the oxalate, tartrate, and fluoride of calcium 
are less soluble than the barium salts. Since borates and tartrates dis¬ 
solve largely in the presence of ammonium salts, a test for H 3 B0 3 is 
always made with the original substance; if indications of organic matter 
have been obtained in the preliminary tests, a special test for tartaric 
acid should be made with the prepared solution. 

Non-formation of a precipitate with BaCl 2 and CaCl 2 
in neutral solution proves the absence of appreciable amounts 
of anions of the first three groups (except oxalate, tartrate, 
and fluoride, as stated above). Acidify the mixture with 
acetic acid. If the precipitate entirely dissolves, acids of 
Groups I and II are absent. If it does not dissolve, add HC 1 
If the precipitate now dissolves acids of Group I are absent. 


ANALYSIS OF A SOLID SUBSTANCE 


145 


and acids of both groups II and III may be present. If the 
precipitate does not dissolve, acids of all three groups may 
be present. 

The precipitate insoluble in HC 1 consists of BaS 0 4 and is taken as a 
test for H 2 S 0 4 , except when the more rare fluosilicate is present, when 
a special test for that anion may be made. In such a case the sulphate 
could be confirmed by the Hepar reaction. 

It would seem by the above procedure that it might be necessary 
to test for all the members of the first three groups in case a precipitate 
insoluble in HC 1 is obtained. But this is not always so. The presence 
of arsenic and chromium will have been made known in the analysis 
for the metals, and tests here only need to be made if they are present, 
to establish the character of the compounds containing them. Pre¬ 
liminary tests have indicated the presence of sulphite, thiosulphate, 
tartrate, carbonate, and fluoride. 

The color of all the barium and calcium salts of the first three groups 
is white, except the chromate, which is yellow. 

(i b ) To a small portion of the prepared solution, add 
HNO3 and AgN03- If no precipitate is formed, acids of 
Group IV are absent. 

The silver salts of Group IV have the following colors: ferricyanide, 
reddish-orange; bromide, yellowish-white; iodide, yellow; sulphide, 
black; the remainder are white. Sulphide will not be present, since 
Et>S will have been removed by boiling the acidified carbonate solution 
to expel C 0 2 . The color of the precipitate may exclude the presence 
of certain acids, while preliminary tests will have indicated the pres¬ 
ence of others of this group. 

(c) Make special tests according to methods described on 
pages 100-126 for acids indicated and not excluded by the above 
group reactions. Note that HNO3 must be tested for in the 
original substance or in a solution prepared without the addi¬ 
tion of HNO3. 

(1 d ) The aqueous solution of the Na2C03 fusion of a sub¬ 
stance, undissolved by dilute acids, prepared according to 
116 ( e ), should be specially tested as below for H2F2 and 


146 


QUALITATIVE CHEMICAL ANALYSIS 


H 3 BO 3 if silicates are present (since the original substance 
would often be undecomposed by the action of cone. H 2 S0 4 ). 
Acidify part of the solution with acetic acid, filter if necessary, 
and add CaCl 2 . Test a precipitate of CaF2 (which may form 
only slowly) for H 2 F 2 by the etching test or by mixing with 
Si0 2 and testing for SiF 4 (p. 108). Evaporate part of the 
aqueous solution of the fusion to dryness, and test for 
H 3 BO 3 with cone. H 2 S0 4 and alcohol. 

Since sulphides are partially oxidized to sulphates by fusion 
with Na 2 CC> 3 , to test for H 2 S0 4 in the presence of H 2 S, it is 
better to boil the original substance with a saturated solution 
of Na 2 CC> 3 . Filter such a mixture and test for sulphate by 
adding HC1 and BaCl 2 . 


APPENDIX 


TABLES OUTLINING THE PROCEDURE FOR THE ANALYSIS 
FOR THE BASE-FORMING AND ACID-FORMING CON¬ 
STITUENTS 1 

I. Analysis for the Base-Forming Constituents 


TABLE I 

ANALYSIS OF GROUP I. 

Treat solution for the analysis for the bases with HC 1 . Filter and wash with 
cold water. Res.: Hg 2 Cl 2 , PbCl 2 , AgCl. Treat with hot water. 


Residue: Hg 2 Cl2, AgCl. 
Treat with NH 4 OH. 


Filtrate: PbCl 2 . Add H 2 S 0 4 . 
Ppt.: white PbS 0 4 . Filter and 
wash; treat residue with NH 4 Ac. 


Residue: Black 
Hg + HgNH 2 Cl. 
Presence of mer¬ 
curous Hg. 


Filtrate: Ag(NH 3 ) 2 Cl. 
Make acid with HN 0 3 . 
Ppt: White AgCl. 
Presence of Ag. 


Filtrate: acetate of lead. Add 
K 2 Cr 0 4 . Ppt.: yellow PbCr 0 4 . 
Presence of Pb. 


1 The student should understand that these tables are only outlines pre¬ 
sented so that the general scheme of analysis may be readily followed. For 
details of the analysis, the preceding pages should be consulted. 


147 








J 48 


QUALITATIVE CHEMICAL ANALYSIS 


TABLE II 

ANALYSIS OF GROUP II, A AND B 

Pass H 2 S into filtrate from Group I. Ppt.: HgS, PbS, Bi 2 S3, CuS, CdS, As 2 S3, 
AS2S5, Sb 2 S 3 , Sb 2 S 6 , SnS, SnS 2 . Filter and wash. Reserve filtrate for Groups III, 
IV, V. Treat res. with (NH 4 ) 2 Sz and filter. 


Res. A, Copper Group. : HgS, PbS, Bi 2 S3, CuS, CdS. Warm with dil. HN 0 3 . 
Filter. 


Res.: HgS. Treat with 
HC 1 and KC 10 S . Boil 
out Cb. Dilute. Add 
SnCl 2 . White Ppt.: 
Hg 2 Cl 2 . Black Ppt.: 
Hg. Presence of mer¬ 

Filt.: Pb(N 0 3 ) 2 , Bi(N 0 3 ) 3 , Cu(N 0 3 ) 2 , Cd(N 0 3 ) 2 . Add 
cone. H 2 S 0 4 , evaporate, dilute, filter. 

Res.: White 

PbS 0 4 . Con¬ 
firm presence 
of Pb as in 
Group I. 

Filt.: Bi 2 (S 0 4 ) 3 , CuS 0 4 , CdS 0 4 . Add 
NH 4 OH and filter. 

curic Hg. 

Res.: Bi(OH) 3 . 
Add Na 2 Sn 0 2 . 
Black ppt.: Bi 
Presence of Bi. 

Filt.: Cu(NH 3 ) 4 S 0 4 , 
Cd(NH 3 ) 4 S 0 4 . 

♦Blue color. Pres¬ 
ence of Cu. Add 
KCN and H 2 S. 
Yellow ppt.: CdS. 
Presence of Cd. 


Filt. B, Arsenic Broup: (NH 4 )3AsS 4 , (NH 4 ) 3 SbS 4 , (NH 4 ) 2 SnS3. Acidify with 
HC 1 and filter. Res.: As 2 S 5 , Sb 2 S 6 , SnS 2 . 

Method I. Warm with cone. HC 1 , dilute, filter. 


Res.: As 2 Ss. Treat 
with HC 1 and KCIO3. 
Add NH 4 OH and 
magnesia mixture. 
Ppt.: MgNH 4 As 0 4 . 

Filt.: SbCl 3 , SnCL 

1. Dilute; divide into two portions. 

To one portion 
add Pt foil and 
Sn. Black stain 
on Pt. Presence 

of Sb. 

To the second portion add Fe, and 
filter. 

Dissolve in HC 1 . Pass 
in H 2 S. Yellow ppt.: 
As 2 S3 and As 2 Ss. 
Presence of As. 

Res.: Sb. 

Dissolve in Tar¬ 
taric acid and 
HNO3. 

Pass in H 2 S. 
Orange ppt.: 
Sb^. Presence 
of Sb. 

Filt.: SnCl 2 . 
Dilute and add 
HgCl 2 . White 
ppt.: Hg 2 Cl 2 . 
Presence of Sn. 




















APPENDIX 


149 


TABLE II ( Continued ) 


Method II for Arsenic Group. Dissolve res. of As*S t , Sb 2 S„, and Sn& in cone. 
HC1 and KC 10 3 . Remove S. Place solution in generator of Marsh apparatus, 
with Zn, H,S 0 4 , and Pt. Pass gases into AgN0 3 . 


Res.: Sn, and black 
deposit of Sb on Pt. 
Dissolve Sn in HC 1 , 
dilute, and test for 
Sn with HgCl 2 . 


Filter AgN 0 3 solution and wash residue. Note metallic 
deposits in the heated tube. 


Res.: SbAg 3 . 

dilute, filter. 
Orange ppt.: 
of Sb. 


Dissolve in HC 1 , 
Add H 2 S. 

Sb 2 S 3 . Presence 


Filt.: H 3 As 0 3 . Add HC 1 
and filter from AgCl. 
To filtrate add H 2 S. 
Yellow ppt.: As 2 S 3 . 
Presence of As. 











150 


QUALITATIVE CHEMICAL ANALYSIS 


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APPENDIX 


151 


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152 


QUALITATIVE CHEMICAL ANALYSIS 


TABLE IV 

ANALYSIS OF GROUP IV. 


To the filtrate from Group III add NH4OH and (NH 4 ) 2 C 0 3 and filter. 


Res.: BaC0 3 , SrC0 3 , CaC0 3 . Ana¬ 
lyze by Method I or II. 

Filt.: To remove traces of barium and 
calcium add (NH 4 ) 2 S 04 and (NH 4 ) 2 C 2 0 4 . 
Filter. To the filtrate add (NH 4 ) 2 HP0 4 ; 
stir, allow to stand, filter. 

Method I. Dissolve in dilute HN0 3 , 
evaporate to dryness, and dehy 1 
drate. Add alcohol-ether mixture, 
filter and wash. 

Res.: White crystalline 
MgNH 4 P0 4 . Pres¬ 
ence of Mg. 

Filt.: Reserve 
for Group V. 

Res.: Ba(N0 3 ) 2 , Sr(N0 3 ) 2 . Dissolve in 50 c.c. 
water, add HAc and K 2 Cr0 4 first to a small 
portion, and if a ppt. forms, to the entire solu¬ 
tion. Boil, filter. 

Filt.: Ca(N0 3 ) 2 . Add H 2 S0 4 . 
White ppt.: CaS0 4 . Presence 
of Ca. Filter, wash res. with 
alcohol, dissolve the CaS0 4 in 
water, add (NH 4 ) 2 C 2 0 4 . White 
ppt.: CaC 2 0 4 . Presence of Ca. 
Confirm with flame test. 

Res.: Yellow BaCr0 4 . 
Presence of Ba. 
Confirm with flame 
test. 

Filt.: Add NH 4 OH 
and (NH 4 ) 2 C0 3 . • 
White ppt.: SrC0 3 . 
Presence of Sr. 
Confirm with flame 
test. 


Method II. Dissolve the residue of BaC0 3 , SrC0 3 , and CaC0 3 in HAc. Dilute, 
and if K 2 Cr0 4 gives a ppt. with a small portion of the solution, add IC 2 Cr0 4 to 
the entire solution, heated to boiling. Filter. 


Res. : Yellow 
BaCr0 4 . Pres- 
e n c e of Ba. 
Confirm with 
flame test. 


Filt.: Divide into two portions. 


To one portion add CaS0 4 , and boil 
several minutes. White ppt.: 
SrS0 4 . Presence of Sr. Confirm 
with flame test. 


To the other portion 
add (NH 4 ) 2 S0 4 , boil, 
filter from SrS0 4 , and 
to the filt., add 
(NH 4 ) 2 C 2 0 4 . White 
ppt.: CaC 2 0 4 . Pres¬ 
ence of Ca. Confirm 
with flame test. 


















APPENDIX 


153 


TABLE V 

ANALYSIS OF GROUP V 


Evaporate filtrate from Group IV to dryness and volatilize ammonium salts 
by ignition. Add 3 c.c. cone. HC1, evaporate, and again ignite. Add 5 c.c. 
cone. HC1 and evaporate just to dryness. Dissolve residue in a very little water 
and filter if necessary. Apply flame test to solution. Yellow flame, presence of 
Na. Violet flame through cobalt glass, presence of K. Divide the solution into 
two portions. 


To one portion add a few drops of HCIO 4 . 
White ppt.: KC10 4 . Presence of K. 
Confirm by flame test. 


To the other portion add an equal 
volume of K 2 H 2 Sb 207 . Stir, allow 
to stand. White, crystalline ppt.: 
Na 2 H 2 Sb 2 (> 7 . Presence of Na. Fil¬ 
ter, dissolve res. in HC1, and con¬ 
firm Na by flame test. 


To the original substance, if in solution add CaO, if a solid add CaO and water. 
Warm, and test vapor by odor and effect on litmus for the presence of NH 4 . 

/ 


II. Analysis for the Acid-Forming Constituents 


TABLE VI 


(Characteristic tests are given for each acid; for other tests, and procedures 
when certain anions interfere, consult the preceding pages.) 

The barium salts of Group I are insoluble in HC1. 

Sulphuric acid, H 2 S0 4 .In the absence of fluosilicic acid, 

the insolubility of BaS0 4 in HC1 
is characteristic. 

Fluosilicic acid, H 2 SiF 6 .Evolves H 2 F 2 with cone. H 2 S0 4 . 


The barium or calcium salts of Group II are soluble in HC1, insoluble in 
HAc. 

Chromic acid, H 2 Cr 04 .Yellow color of BaCr 04 , and other 

indications. 

Oxalic acid, H 2 G 2 0 4 .Evolves C0 2 and CO with cone. 

H 2 S0 4 . 

Sulphurous acid, H 2 S 03 .Odor of S0 2 with IUSO 4 . 

Thiosulphuric acid, H 2 S 2 0 3 .Evolves S0 2 and deposits S with cone. 

H 2 S0 4 . 

Hydrofluoric acid, H 2 F 2 .Etching test 













154 


QUALITATIVE CHEMICAL ANALYSIS 


TABLE VI ( Continued ) 


The barium or calcium salts of Group III are soluble in HC1 and HAc, insolu¬ 
ble in neutral solution. 

Phosphoric acid, H3PO4 .Yellow ppt. with ammonium mo¬ 

lybdate. 

Arsenious acid, H 3 As0 3 .With HC1 and H 2 S, arsenic sulphide 

precipitates quickly. 

Arsenic acid, H 3 As0 4 .With HC1 and H 2 S, arsenic sulphide 

precipitates slowly. 

Boric acid, H 3 B0 3 .Turmeric paper; green flame with 

cone. H 2 S0 4 and alcohol. 

Silicic acid.Bead test. Test with H 2 F 2 . 

Carbonic acid, H 2 C0 3 .Evolves C0 2 with acids. 

Tartaric acid, H 2 C 4 H 40 6 .The acid potassium salt is insoluble. 


The silver salts of Group IV are insoluble in HN0 3 . 

Hydrochloric acid, HC1.In the presence of other halogen 

acids, test with Pb0 2 . 

Hydrobromic acid, HBr.. .Bromine is liberated by the action of 

chlorine. 

Hydriodic Acid, HI.Iodine is liberated by the action of 

chlorine. 

Hydrocyanic acid, HCN.Forms Prussian blue with iron salts. 

Hydrogen sulphide, H 2 S.Blackens lead acetate paper. 

Hydroferrocyanic acid, H 4 Fe(CN) 6 .Prussian blue with ferric salts. 

Hydroferricyanic acid, H 3 Fe(CN) 4 .Turnbull’s blue with ferrous salts. 

Sulphocyanic acid, HSCN.Red color with ferric salts. 


The barium and silver salts of Group V are soluble in water. 


Nitric acid, HN0 3 .Brown ring test with cone. H 2 S0 4 

and FeS0 4 . 

Nitrous acid, HN0 2 .Oxidizes iodides to iodine. 

Acetic acid, HC 2 H 3 0 2 .Odor of ethyl acetate with cone. 

H 2 S0 4 and alcohol. 

Chloric acid, HC10 3 .Evolves explosive CIO 2 with cone. 

h 2 sc 4 . 
























APPENDIX 


155 


TABLE VII 

Directions for the Preparation of Reagents 
Test Solutions and Liquid Reagents 

(The name of the substance is followed by its molecular formula, the normality 

of the solution, and in parenthesis the number of grams of the substance to be dis¬ 
solved in a liter of solution, or any special directions necessary.) 

Acetic acid, HC 2 H 3 0 2 ,6-N. (350 c.c. glacial acetic acid in 1 liter.) 

Alcohol, CzHeO (95%). 

Alcohol-ether. (Mixture of equal parts absolute alcohol and dry ether.) 

Aluminium nitrate, A 1 (N 0 3 ) • 7 • 5H2O, N/2. (58.) 

Ammonium acetate, NH 4 Ac, 3-N. (Add 500 c.c. 6-N HAc to 500 c.c. 6-N 
NH4OH). m 

Ammonium carbonate, (NH 4 ) 2 C 0 3 . (Dissolve 250 g. ammonium carbonate in 
750 c.c. 6-N NH 4 OH, filter if necessary, and add 250 
c.c. 15-N NH 4 OH). 

Ammonium chloride, NH 4 C 1 , i-N. (53.5.) 

Ammonium hydroxide, NILOH, 15-N. (Cone., sp.g. 0.90.) 

Ammonium hydroxide, NH 4 OH, 6-N. (Add 400 c.c. 15-N NH 4 OII to 600 c.c. 
water.) 

Ammonium molybdate solution. (Dissolve 75 g. ammonium molybdate in 500 c.c. 

water, and add to 500 c.c. 6-N HN 0 3 .) 

Ammonium nitrate, NH1NO3, N/2. (40.) 

Ammonium oxalate, (NH 4 ) 2 C 2 0 4 *H 2 0 , N/2. (35.5*) 

Ammonium phosphate, (NH 4 )2HP0 4 , i-N. (116.) 

Ammonium polysulphide (yellow), (NH 4 ) 2 Sz. (Heat colorless ammonium sul¬ 
phide with flowers of sulphur for several hours, and 
filter.) 

Ammonium sulphate, (NH^SCh, N/2. (33.) 

Ammonium sulphide (colorless), (NH 4 ) 2 S. (Saturate 300 c.c. 6-N NH 4 OH with 
H 2 S, add 200 c.c. 6-N NH 4 OH and dilute to a liter.) 

Ammonium sulphocyanate, NH 4 SCN, N/ 2. (38.) 

Antimony pentachloride, SbCl 5 , N/2. (29.7 g. to a liter, HC 1 being added to dis¬ 

solve basic salts.) 

Antimony trichloride, SbCl 3 , N/2. (37.7 g. to a liter, HCl being added to dissovle 

basic salts.) 

Arsenious acid. (8.2 g. As 2 0 3 in a liter.) 

Barium chloride, BaCl 2 ’2H 2 0, i/N. (122.) 

Barium hydroxide, Ba( 0 H) 2 - 8 H 2 0 , N/5. (31.6.) 

Barium nitrate, Ba(N 0 3 ) 2 , N/2. (65.3.) 

Bismuth nitrate, Bi(N 0 3 ) 3 * 5 H 2 0 , N/* (40.3 g. to a liter, HN 0 3 being added to 

dissolve basic salts.) 

Bromine, saturated solution in water. 

Brucine. (0.2 g. in 100 c.c. cone. H 2 S 0 4 .) 


156 


QUALITATIVE CHEMICAL ANALYSIS 

TABLE VII (< Continued ) 

Cadmium nitrate, Cd(N03)2*4H 2 0, N/2. (77.) 

Calcium chloride, CaCl 2 * 6 H 2 0 , i-N. (109.5.) 

Calcium hydroxide, Ca(OH) 2 . (Saturated solution.) 

Calcium nitrate, Ca(N 0 3 ) 2 - 2H2O, N/2. (59.) 

Calcium sulphate, CaS 0 4 • 2II2O. (Saturated solution.) 

Carbon bisulphide, CS 2 . 

Chlorine, saturated solution. 

Chromium nitrate, Cr(N 0 3 ) 3 , N/2. (39.7*) 

Cobalt nitrate, Co(N 0 3 ) 2 * 6 H 2 0 , N/2. (72.8.) 

Cobalt sulphate, CoS 0 4 - 7H2O, N/2. (70.) 

Cupric nitrate, Cu(N0 3 ) 2 *3H 2 0, N/2. (60.4.) 

Dimethyl-glyoxime, C 4 HgN 2 0 2 . (1 per cent solution in 95 per cent alcohol.) 

Ferric chloride, FeCl 3 * 6 H 2 0 , N/2. (45.) 

Ferric nitrate, Fe(N 0 3 ) 3 *oH 2 0 , N/2. (67.4.) 

Hydrochloric acid, HC 1 , 13-N. (Cone., sp. g. 1.20.) 

Hydrochloric acid, HC 1 , 6-N. (Dilute 465 c.c. 13-N HC 1 to a liter.) 
Hydrochlorplatinic acid, H 2 PtCl 6 - 6 H 2 0 , N/5. (5.2 g. in 100 c.c.) 

Hydrofluoric acid, H 2 F 2 . (The 48 per cent solution on the market is used un¬ 
diluted.) 

Hydrogen peroxide. (3 per cent solution.) 

Lead acetate, PbAc 2 *3H 2 0, N/2. (94.7.) 

Lead nitrate, Pb(N 0 3 ) 2 , N/2. (82.7.) 

Magnesia mixture. (Dissolve 101 g. magnesium chloride and 100 g. ammonium 
chloride in water, add 60 c.c. cone. NH 4 OH, and dilute 
to a liter.) 

Magnesium nitrate, Mg(N 0 3 ) 2 - 6 H 2 0 , N/2. (64.1). 

Manganous nitrate, Mn(N 0 3 ) 2 - 6 H 2 0 , N/2. (71.8.) 

Mercuric chloride, HgCl 2 , N/2. (67.7.) 

Mercuric nitrate, Hg(N 0 3 ) 2 , N/2. (81.) 

Mercurous nitrate, HgN 0 3 *H 2 0 , N/2. (140-) 

Nessler’s reagent. (Dissolve 12 g. HgCl 2 in 100 c.c. water and add a solution of 
14.8 g. KI in 100 c.c. water; allow to settle and wash 
by decantation several times; add a solution of 10 g. 
KI which will dissolve the residue; add 50 c.c. 6-N 
NaOH, dilute to 250 c.c., allow to settle, and decant. 
Keep the solution in the dark.) 

Nickel nitrate f Ni(N0 3 )2*6H 2 0, N/2. (72.7.) 

Nitric acid, HN 0 3 , 16-N. (Cone., sp.g. 1.42.) 

Nitric acid, HN 0 3 , 6-N. (Dilute 382 c.c. cone, acid to a liter.) 

Nitroso beta naphthol, C10H7NO2. (1 per cent solution in 6-N acetic acid.) 

Perchloric acid, HC10 4 . (Use the solution on the market, undiluted.) 

Potassium arseniate, K 2 HAs 0 4 . (22 g. to a liter.) 

Potassium arsenite, K 2 HAs 0 3 . (20 g. to a liter.) 


APPENDIX 


157 


TABLE VII (< Continued ) 

Potassium chloride, KC 1 , i-N. (75.6.) 

Potassium chromate, K 2 Cr 0 4 , N/2. (48.6.) 

Potassium cyanide, KCN, i-N. (65.2.) 

Potassium ferricyanide, K 3 Fe(CN) 6 , i-N. (109.8.) 

Potassium ferrocyanide, K 4 Fe(CN) 6 *3H20, i-N. (105.7.) 

Potassium iodide, KI, N/5. (33.2.) 

Potassium nitrite, KNO 2 , 2-N. (170.2.) 

Potassium pyroantimonate, K 2 H 2 Sb20 7 . (Boil 25 g. of the salt in a liter of water, 
cool the solution, and add a solution of 5 g. KOH in 
50 c.c. water. 

Potassium sulphocyanate, KSCN, N/2. (58.6.) 

Silver nitrate, AgN 0 3 , N/5. (34.) 

Silver sulphate, Ag 2 S 0 4 . (Saturated solution.) 

Sodium acetate, NaAc^FLO, i-N. (136.) 

Sodium carbonate, Na 2 C 0 3 , i-N. (53.) 

Sodium chloride, NaCl, i-N. (58.5.) 

Sodium cobaltinitrite, Na 3 Co(N 0 2 ) 6 . (Dissolve 125 c.c. g. sodium nitrite in 200 
c.c. water and add 65 c.c. 6-N acetic acid and a solu¬ 
tion of 13 g. cobalt nitrate in 235 c.c. water. Filter 
before use.) 

Sodium hydroxide, NaOH, 6-N. (245.) 

Sodium phosphate, Na 2 HP0 4 - 12 H 2 O, i-N. (119.4.) 

Sodium polysulphide, Na 2 Sz. (Saturate 300 c.'c. 6-N NaOH with H 2 S, add 200 
c.c 6-N NaOH and dilute to a liter.. Heat this solu¬ 
tion with flowers of sulphur for several hours and filter.) 

Sodium silicate, Na 2 Si0 3 *xH20. Commercial solution of “ water-glass.” 

Stannic chloride, SnCh, N/2. (32.6.) 

Stannous chloride, SnCl 2 *2H20, N/2. (Dissolve 56.5 g. of the salt in water, add¬ 
ing enough cone. HC 1 to clear the solution, dilute to a 
liter, and keep granulated tin in the bottle.) 

Starch solution. (Dissolve 1 g. soluble starch in 250 c.c. boiling water, and add a 
little chloroform to preserve the solution.) 

Strontium nitrate, Sr(N 0 3 ) 2 , N/2. (52.9.) 

Sulphuric acid, H 2 S 0 4 . (Cone., sp. g. 1.84.) 

Sulphuric acid, H2S0 4 , 6-N. (Add 167 c.c. cone, acid to 833 c.c. water.) 

Zinc nitrate, Zn(N0 3 )2’6H 2 0, N/2. ( 744 -) 

DRY REAGENTS 

Aluminium sulphate, A1 2 (S0 4 ) 3 * i8H 2 0. 

Ammonium chloride, NH 4 C 1 . 

Ammonium dichromate, (NH 4 )2Cr 2 0 7 . 

Ammonium fluoride, (NHO2F2. 



158 


QUALITATIVE CHEMICAL ANALYSIS 


TABLE VII ( Continued) 

Ammonium oxalate, (NH 4 )2C20 4 -2H 2 0. 

Arsenic trioxide, AS2O3. 

Barium carbonate, BaC03. 

Barium hydroxide, Ba(0H) 2 ’8H20. 

Bismuth nitrate, Bi(N0 3 )3. 

Borax, Na2B 4 0 7 - ioH 2 0. 

Cadmium sulphate, CdS0 4 -4H 2 0. 

Calcium carbonate, CaCCb. 

Calcium fluoride, CaF2. 

Calcium oxide, CaO. 

Charcoal sticks. 

Chrome-alum, KCr(S0 4 )2- 12H2O. 

Cobalt nitrate, Co(N0 3 )2-6H 2 0. 

Cupric chloride, CuCl2*2H 2 0. 

Ferrous sulphate, FeS 0 4 - 7H2O. 

Ferrous sulphide, FeS. 

Iodine. 

Iron wire (pure). 

Lead dioxide, PbCh. 

Litmus paper, red and blue. 

Manganese sulphate, MnS0 4 *7H 2 0. 

Mercuric chloride, HgCh. 

Mercuric nitrate, Hg(N0 3 )2- 
Mercurous chloride, Hg 2 Cl2. 

Methyl violet paper. Immerse filter paper in a 0.1 per cent solution of methyl 
violet (Kahlbaum’s Krystallviolett) and dry. 

Nickel sulphate, NiS 0 4 - 6 H 2 0 . 

Paraffin. 

Platinum foil and wire. 

Potassium acid sulphate, KHS 0 4 . 

Potassium antimonyl tartrate, K(Sb 0 )C 4 H 4 06 . 

Potassium bromide, KBr. 

Potassium carbonate, K2CO3. 

Potassium chlorate, KCIO3. 

Potassium chloride, KC 1 . 

Potassium cyanide, KCN. 

Potassium dichromate, K 2 Cr 2 07. 

Potassium iodide, KI. 

Potassium nitrate, KNO3. 

Potassium nitrite, KNO2. 

Sand. 

Sodium acetate, NaC 2 H 3 02-3H 2 0. 

Sodium ammonium hydrogen phosphate, NaNH 4 HP0 4 *4H 2 0. 


APPENDIX 


159 


TABLE VII ( Continued) 


Sodium carbonate, Na2CC>3. 

Sodium chloride, NaCl. 

Sodium nitrate, NaN 0 3 . 

Sodium nitrite, NaN02. 

Sodium peroxide, Na 2 0 2 . 

Sodium potassium tartrate, NaK^ILOe^ILO. 
Strontium nitrate, Sr(N 0 3 ) 2 . 

Tartaric acid, H2C4H4O6. 

Tin, granulated. 

Turmeric paper. 

Zinc, granulated. 

Zinc sulphate, ZnS04*7H 2 0. 


160 


QUALITATIVE CHEMICAL ANALYSIS 


TABLE VIII 

Solubilities of Difficultly Soluble Substances in Water 


(Solubilities are expressed in milligrams of anhydrous substance per liter of 
solution.) 


Substance. 

Temp. 

Degrees 

Cent. 

Milligrams 
• per Liter. 

Substance. 

Temp. 

Degrees 

Cent. 

Milligrams 
per Liter. 

AgBr 

25 

0.14 

CuS 

25 

O.O 15 I 

AgCl 

25 

1.8 

Hgl 2 

18 

0.4 

Ag 2 Cr 0 4 

18 

25 - 

HgS 

25 

0.0^4 

Agl 

25 

0.003 

KC 10 4 

25 

20,000. 

Ag 3 P 0 4 

20 

6. 

K 2 PtCl 6 

20 

11,000. 

Ag 2 S 

25 

0. On6 

MgF 2 

l8 

80. 

Ag 2 S 0 4 

25 

8,000. 

MgC 2 0 4 

l8- 

300. 

AgSCN 

25 

0.2 

Mg( 0 H) 2 

18 

9 - 

BaCOs 

25 

20. 

PbBr 2 

25 

9,700. 

BaC 2 0 4 

18 

90. 

PbCl 2 

25 

11,000. 

BaCr 0 4 

18 

4 - 

PbCr 0 4 

18 

0.2 

BaF 2 

18 

1,600. 

Pbl 2 

25 

760. 

Ba(OH) 2 

18 

37,000. 

PbS 

25 

0.012 

BaS 0 4 

26 

2.6 

PbS 0 4 

18 

41. 

CaC 0 3 

20 

12. 

SrC 0 3 

25 

7 - 

CaC 2 0 4 

18 

5-5 

SrC 2 0 4 

18 

46. 

CaCr 0 4 

18 

4,000. 

SrCr 0 4 

18 

1,200. 

CaF 2 

18 

16. 

SrF 2 

18 

117. 

Ca(OH) 2 

18 

1,700. 

Sr( 0 H) 2 

18 

8,000. 

CaS 0 4 

25 

2,000 

SrS 0 4 

18 

118. 

















APPENDIX 


161 


TABLE IX 

Table of Solubilities 1 

Showing the classes to which the compounds of the commonly occurring elements 
belong in respect to their solubility in water , hydrochloric acid , nitric acid , or 
aqua regia. 

Preliminary Remarks 

For the sake of brevity, the classes to which the compounds belong are expressed 
in letters. These have the following signification: 

W or w, soluble in water. 

A or a, insoluble in water, but soluble in hydrochloric acid, nitric acid, or in 
aqua regia. 

I or i, insoluble in water, hydrochloric acid, or nitric acid. 

Further, substances standing on the border-lines are indicated as follows: 

W-A or w-a, difficultly soluble in water, but soluble in hydrochloric acid or 
nitric acid. 

W-I or w-i, difficultly soluble in water, the solubility not being greatly in¬ 
creased by the addition of acids. 

A-I or a-i, insoluble in water, difficultly soluble in acids. 

If the behavior of a compound to hydrochloric and nitric acids is essentially 
different, this is stated in the notes. 

Capital letters indicate common substances used in the arts and in medicine, 
while the small letters are used for those less commonly occurring. 

The salts are generally considered as normal, but basic and acid salts, as 
well as double salts, in case they are important in medicine or in the arts, are 
referred to in the notes. 

The small numbers in the table refer to the following notes. 

1 Fresenius’ Qualitative Analysis, Wells’ translation of i6th German Edition. 


162 


QUALITATIVE CHEMICAL ANALYSIS 


SOLUBILITY 



Potassium. 

Sodium. 

Ammonium. 

Barium. 

Strontium. 

» 

E 

3 

l 

Magnesium. 

Aluminium 

Chromium. 

£ 

Manganese. 

Nickel. 

Cobalt. 

Oxide. 

w 

w 

w 

w 

W 

W-A 

A 

A 

A & I 

A 

ai7 

A 

A 

Chromate. 

Wi 

w 

w 

a 

w-a 

w-a 

w 


a 

w 

w 

a 

a 

Sulphate. 

W,3.15 

W 

W 14.20-30 

I 

I 

W-I 

W 

W,3.1 4 

W&I 15 

W 

W 

W 

W 

Phosphate. 

w 

Ws 

W 8 .12 

a 

a 

An 

ai2 

a 

a 

a 

a 

a 

a 

Borate. 

W 2 

w 9 

W 

a 

a 

a 

w-a 

a 

a 

a 

a 

a 

a 

Oxalate.. 

w 3 

w 

w 

a 

a 

A 

a 

a 

w-a 

a 

w-a 

a 

a 

Fluoride. 

w 

w 

W 

w-a 

w-a 

A-I 

a-i 

w 

w 

w-a 

a 

w-a 

w-a 

Carbonate. 

W 4 

W 10 

W 

A 

A 

A 

A 



A 

A 

A 

A 

Silicate. 

W 

w 


a 

a 

a 

a 

a-i 

a 

a 

a 

a 

a 

Chloride. 

W 37 

w 35 

W21.38 

W 

W 

W 

W 

w 

W&I 

W 

W 

W 

W 

Bromide. 

W 

w 

w 

w 

w 

w 

w 

w 

w & i 

w 

w 

w 

w 

Iodide . 

w 

w 

W 

w 

w 

w 

w 

w 

w 

w 

w 

w 

w 

Cyanide . 

w 

w 

w 

w-a 

w 

w 

w 


a 

A 

a 

a-i 

a-i 

Ferrocyanide.... 

w 

w 

w 

w-a 

w 

w 

w 



A-I 

a 

i 

i 

Ferricyanide.... 

w 

w 

w 



w 

w 



a 

i 

i 

i 

Sulphocyanate.. 

w 

w 

W 

w 

w 

w 

w 


w 

w 

w 

w 

w 

Sulphide. 

w 

W 

W 

W 

w 

W-A 45 

a 

a 

a-i 

Ak, 

A 


an* 

Nitrate. 

w 

W 

W 

W 

W 

w 

w 

w 

W 

w 

w 

w 

W 

Chlorate. 

w 

w 

w 

w 

W 

w 

w 

to 

w 

w 

w 

w 

w 

Tartrate. 

W5.0.7-22-40 

W 7 

w 6 

a 

a 

A 

w-a 

w 

w 

a 

w-a 

a 

w 

Citrate. 

w 

w 

w 

a 

a 

w-a 

w 

w 

w 

w-a 

a 

w 

w 

Malate. 

w 

w 

w 

w & a 

w 

w-a 47 

w 

w 


w 

w 



Succinate. 

w 

w 

w 

w-a 

w-a 

w-a 

w 

w-a 


w-a 

w 

w 

w-a 

Benzoate. 

w 

w 

w 

w 


w 

w 




w 



Salicylate. 

w 

W 

W 

w-a 

w-a 

w-a 

w 







Acetate. 

w 

W 

W 

W 

w 

W 

w 

W 

to 

W 

w 

w 

w 

Formate. 

w 

w 

w 

w 

w 

w 

w 

w 

w 

w 

w 

w 

w 

Arsenite. 

w 

w 

'W 

a 

a 

a 

a 




a 

a 

a 

Arsenate. 

w 

W 

w 

a 

a 

a 

a 

a 

a 

a 

a 

a 

a 





















































APPENDIX 


163 


TABLE 


Ferrous 

Ferric. 

Silver. 

• 

Lead. 

i 

3 

g 

s 

Mercuric. 

Cupric. 

Bismuth. 

Cadmium. 

Gold. 

Platinum. 

Stannous. 

Stannic. 

Antimonous. 


a 

A 

a 

A 2 4 

A 

A 

A 

a 

a 


a 

a 

a &i 

A4 2 

Oxide 


w 

a 

A-I 

a 

w-a 

w 

a 

a 



a 


a 

Chromate 

w 20 

W 

W-A 

A-I 

w-a 

W 27 

w 30 

w 

w 


w 

w 


a 

Sulphate 

a 

A 

a 

a 

a 

a 

a 

a 

a 



a 

a 

w-a 

Phosphate 

a 

a 

a 

a 



a 

a 

w-a 



a 



Borate 

a 

a 

a 

a 

a 

a 

a 

a 

a 


w 

a 

w 

.•a 

Oxalate 

w-a 

• w 

w 

a 


w-a 

a 

w 

w-a 



w 

w 

w 

Fluoride 

A 


a 

A 

a 

a 

A 

a 

a 






Carbonate 

a 

a 


a 



a 


a 






Silicate 

W 

w 21 

I 

W-I 

A-I 

w 28 

W 

W-A 33 

W 

Was 

W 3 7.38 

W 

W 4 o 

W-A 42 

Chloride 

\v 

w 

i 

w-i 

a-i 

w 

w 

w-a 

W 

w 

W 



w-a 

Bromide 

W 

w 

i 

W-A 

A 

A 

w 

a 

W 

a 

i 

w 

w 

w-a 

Iodide 

a-i 


I 

a 


W 

a 


a 

W 

w 




Cyanide 

i 

I 

i 

a 



i 





i 

i 


Ferrocyanide 

I 

w 

i 

w-a 







• 

i 



Ferricyanide 

w 

w 

i 

a 

A 

w 

a 


w-a 


a 


w 


Sulphocyanate 

A 

a 

a 2 3 

A 

A 

A 2 9 

a3i 

a 

A 

a3o 

a39 

a4i 

a4i 

A 4445 

Sulphide 

vv 

w 

W 

W 

W 2C 

w 

W 

W 34 

w 


w 




Nitrate 

w 

w 

w 

w 

w 

w 

w 

w 

w 



w 



Chlorate 

w-a 

W 22 

a 

a 

w-a 

a 

w 

a 

w-a 



a 


a4o. 

Tartrate 

w 

W 

a 

a 

a 

w-a 

w 


a 






Citrate 


W 

w-a 

w-a 

a 

w-a 

w 





w 

w 


Malate 

w-a 

a 

a 

a 

a 

w-a 

w 


w 




a 


Succinate 

w 

a 

w-a 

a 

a 

w-a 

a 


w 






Benzoate 



w-a 

w-a 



w 








Salicylate 

w 

W 

w 

W 25 

w-a 

w 

W 32 

w 

w 



w 

w 


Acetate 

w 

W 

w 

w-a 

w 

w 

w 

w 

w 



w 



Formate 

a 

a 

a 

a 

a 

a 

A 







a 

Arsenite 

a 

a 

a 

a 

a 

a 

a 

a 





a 

a 

Arsenate 




































164 


QUALITATIVE CHEMICAL ANALYSIS 


NOTES TO TABLE OF SOLUBILITIES 

(i) Potassium dichromate, W. (2) Potassium borotartrate, W. (3) Hydro¬ 
gen potassium oxalate. W. (4) Hydrogen potassium carbonate, W. (5) Hydro¬ 
gen potassium tartrate, W. (6) Ammonium potassium tartrate, W. (7) Sodium 
potassium tartrate, W. (8) Ammonium sodium phosphate, W. (9) Acid sodium 
borate, W. (10) Hydrogen sodium carbonate, W. (11) Tricalcium phosphate, 
A. (12) Ammonium magnesium phosphate, A. (13) Potassium .aluminium 
sulphate, W. (14) Ammonium aluminium sulphate, A. (15) Potassium chro¬ 
mium sulphate, W. (16) Zinc sulphide, as sphalerite, soluble in nitric acid with 
separation of sulphur; in hydrochloric acid, only upon heating. (17) Man¬ 
ganese dioxide, easily soluble in hydrochloric acid; insoluble in nitric acid. (18) 
Nickel sulphide is rather easily decomposed by nitric acid; very difficultly by 
hydrochloric acid. (19) Cobalt sulphide, like nickel sulphide. (20) Ammonium 
ferrous sulphate, W. (21) Ammonium ferric chloride, W. (22) Potassium ferric 
tartrate, W. (23) Silver sulphide, only soluble in nitric acid. (24) Minium 
is converted by hydrochloric acid into lead chloride; by nitric acid, into soluble 
lead nitrate and brown lead peroxide which is insoluble in nitric acid. (25) 
Tribasic lead acetate, W. (26) Mercurius solubilis Hahnemanni, A. (27) 
Basic mercuric sulphate, A. (28) Mercuric chloride-amide, A. (29) Mercuric 
sulphide, not soluble in hydrochloric acid, nor in nitric acid, but soluble in aqua 
regia upon heating. (30) Ammonium -cupric sulphate, W. (31) Copper sul¬ 
phide, is decomposed with difficulty by hydrochloric acid, but easily by nitric 
acid. (32) Basic cupric acetate, partially soluble in water, and completely in 
acids. (33) Basic bismuth chloride, A. (34) Basic bismuth nitrate, A. (35) 
Sodium auric chloride, W. (36) Gold sulphide is not dissolved by hydrochloric 
acid, nor by nitric acid, but it is dissolved by hot aqua regia. (37) Potassium 
platinic chloride, W-I. (38) Ammonium platinic chloride, W-I. (39) Plat¬ 
inum sulphide is not attacked by hydrochloric acid, is but slightly attacked by 
boiling nitric acid (if it has been precipitated hot), but is dissolved by hot aqua 
regia. (40) Ammonium stannic chloride, W. (41) Stannous sulphide and stannic 
sulphide are decomposed and dissolved by hot hydrochloric acid, and are con¬ 
verted by nitric acid into oxide which is insoluble in an excess of nitric acid. 
Sublimed stannic sulphide is dissolved only by hot aqua regia. (42) Antimonous 
oxide, soluble in hydrochloric acid, not in nitric acid. (43) Basic antimonous 
chloride, A. (44) Antimony sulphide is completely dissolved by hydrochloric 
acid, especially upon heating; it is decomposed by nitric acid, but dissolved 
only to a slight degree. (45) Calcium antimony sulphide, W-A. (46) Potas¬ 
sium antimony tartrate, W. (47) Hydrogen calcium malate, W. 


APPENDIX 


165 


TABLE X 

Percentage Ionization of Acids, Bases, and Salts 

(Approximate percentage ionization values are given for solutions of o.i nor¬ 
mality at i8°. For the polybasic acids, the entire ionization is assumed to be only 
primary ionization.) 

HBr, HC 1 , HCIO3, HCIO4, HI, HNO3, HMnCh.90% and above 

H2SO4. 51 

H2C2O4.50 

H3PO4.26 

H2F2. 9 

H2C4H4O6. 8 

HC2H3O2. 1 

H2CO3. 0.2 

H2S. 0.07 

HCN. 0.01 

KOH.89 

NaOH.84 

Ba(OH) 2 ...80 

NH4OH. 1 

Sr(OH) 2 (N/6 4 ).93 

Ba(OH) 2 (N/6 4 ).92 

Ca(OH) 2 (N/6 4 ).90 

Salts of the class C +1 A —1 (C= cation; A = anion).86 

Salts of the class C +2 A 2 -1 .72 

Salts of the class C 2 +1 A~ 2 .72 

Salts of the class C +2 A -2 .45 

HgCl 2 , Hg(CN) 2 . 1 
























TABLE XI 

TABLE OF THE ELEMENTS ARRANGED ACCORDING TO THE PERIODIC SYSTEM 

(Many of the rarer elements are not Inserted) 


166 


QUALITATIVE CHEMICAL ANALYSIS 




P 

§ 

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a 


gs 

ll 

ao 

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10*0 

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dd 

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53*0 

aa 


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V- 

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dd 

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05 


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|| 


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N 

rH 




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cc 

a 

a 

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0 



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00 

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CG 

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q 

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Sb= 120. 

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o 

Pi 

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a 





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Rb = 85.45 

oo 




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co 

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Cs= 132.8 

05 

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APPENDIX 


167 


TABLE XII. INTERNATIONAL ATOMIC WEIGHTS, 1920 



Sym¬ 

bol. 

Atomic 

weight. 


Sym¬ 

bol. 

Atomic 

weight. 

Aluminium. 

A 1 

27.1 

Molybdenum. 

Mo 

06.0 

Antimony. 

Sb 

120.2 

Neodymium. 

Nd 

I44.3 

A rgon. 

A 

TO 0 

Neon. 

Ne 

20.2 

Arsenic. . 

As 

74.96 

Nickel. 

Ni 

58.68 

Barium. 

Ba 

137-37 

Niton (radium 


Bismuth 

Bi 

208.0 

emanation). 

Nt 

222.4 

Boron., 

B 

10.9 

Nitrogen. 

N 

14.008 

Bromine, 

Br 

79.92 

Osmium. 

Os 

190.9 

C nrlmiiim 

Cd 

112.40 

Oxygen. 

0 

16.00 

Caesium.. 

Cs 

132.81 

Palladium. 

Pd 

106.7 

Calcium. 

Ca 

40.07 

Phosphorus. 

P 

31.04 

Carbon. . 

c 

12.005 

Platinum. 

Pt 

195.2 

Cerium 

Ce 

140.25 

Potassium. 

K 

39 10 

Chlorine. 

Cl 

35.46 

Praseodymium. 

Pr 

140.9 

Chromium 

Cr 

<2 .O 

Radium. 

Ra 

226.0 

C obalt 

Co 

T8.97 

Rhodium. 

Rh 

102.9 

Colnmhinm 

Cb 

0T • I 

Rubidium. 

Rb 

85.45 

C onner 

Cu 

63 • 37 

Ruthenium. 

Ru 

101 . 7 

Ovsnrosium 

Dv 

162. q 

Samarium. 

Sa 

1504 

Erbium 

Er 

167.7 

Scandium. 

Sc 

44.1 

En roni 11m 

Eu 

IC 2 .O 

Selenium. 

Se 

79.2 

Eluorine 

F 

J 

IQ . O 

Silicon. 

Si 

28.3 

Oannlirmim 

Gd 

7 

i «57 . z 

Silver. 

Ag 

107.88 

Callinm 

Ga 

* j / j 

70.1 

Sodium. 

Na 

23.00 

VJTCtlXlU .111 . 

Cermaninm 

Ge 

72. 3 

Strontium. 

Sr 

87.63 

m nrimim 

G 1 

0 . 1 

Sulphur. 

S 

32.06 

Gold 

Au 

V 

197.2 

Tantalum. 

Ta 

181.5 

Helium 

He 

4.00 

Tellurium. 

Te 

127-5 

Holminm 

Ho 

163 . 5 

Terbium. 

Tb 

I 59 - 2 

TJ\rH morpn 

H 

0 • 0 

1.008 

Thallium . 

Tl. . 

204.0 

Tndium 

In 

114.8 

Thorium . 

Th 

232.15 

Tori i tie 

I 

126.92 

Thulium. 

Tm 

168.5 

T rinnim 

Ir 

IQT . I 

Tin. 

Sn 

118.7 

Iron. 

Fe 

55-84 

Titanium. 

Ti 

48.1 

Y r\mton 

Kr 

82.92 

Tungsten. 

W 

184.0 

T QntnQ rmm 

La 

139.0 

207 . 20 

Uranium . 

U 

238.2 

T.earl 

Pb 

Vanadium . 

V 

510 

T ithinm 

Li 

6 . 04 

V/ . 

Xenon . 

Xe 

130.2 

Lutecium . 

Lu 

175-0 

Ytterbium 



Magnesium . 

Mg 

24.32 

(Neoytterbium) . . 

Yb 

173-5 


Mn 

54-93 

200 . 6 

Yttrium . 

Yt 

89-33 

lvianganese . 

TV/T ot*riii*\r 

Hff 

Zinc . 

Zn 

65-37 




Zirconium . 

Zr 

90.6 

































































































INDEX 


Acetic acid, 124 
Acids, groups of, 99 
Alloys and metals, analysis of, 136 
Aluminium, 71 
Ammonium, 96 
Amphoteric electrolytes, 35 
Analysis, acid, 133, 142 
basic, 27 

of alkali group, 98 
of alkaline earth group, 92 
of arsenic group, 64 
of copper group, 63 
of iron and zinc groups, 82 
of silver group, 42 
qualitative, 1 
quantitative, 1 
Anion, 9 
Antimony, 57 
Aqua regia, 45 
Arsenic, 54 
Arsenic acid, no 
Arsenious acid, 110 
Avogadro’s hypothesis, 3 

Barium, 90 
Bases, groups of, 40 
Bead test, 49 
Bismuth, 46 


Boric acid, m 
Boyle’s Law, 2 

Cadmium, 53 
Calcium 90 
Carbonic acid, 114 
Cation, 9 

Chemical equilibrium, 14 
Chloric acid, 125 
Chromic acid, 102 
Chromium, 72 
Cobalt, 78 
Colloids, 56 
Complex ions, 29 
Constant, equilibrium, 15 
ionization, 17 
hydrolytic, 23 
Conductivity, molecular, 1 
Copper, 47 

Dalton’s law, 3 
Decrepitation, 125 
Deflagration, 125 
Dissociation, 4 
Dry reactions, 33 

Effect, common-ion, 18 
Electrolysis, 9 

169 





170 INDEX 


Electrolyte, 8 
Electrons, 25 
Equilibrium, 14 

Flame tests, 91 

Gay Lussac’s law, 2 

Hepar reaction, 100 
Hydriodic acid, 117 
Hydrobromic acid, 117 
Hydrochloric acid, 117 
Hydrocyanic acid, 119 
Hydroferricyanic acid, 120 
Hydroferrocyanic acid, 120 
Hydrofluoric acid, 107 
Hydrofluosilicic acid, 101 
Hydrogen sulphide, 120 
Hydrolysis, 21 

Indicators, 28 
Introduction, 1 
Ion, 9 

Ionization, 8 
Iron, 68 

Law of mass-action, 13 
Lead, 32 

Magnesium, 90 
Manganese, 75 
Marsh test, 67 
Mercury, 27, 43 
Metaphosphate bead test, 49 

Nessler’s reagent, 46 


Nickel, 80 
Nitric acid, 122 
Nitrous acid, 123 
Normal solution, n 

Organic substances, detection, 128, 
132 

removal of, for basic analysis, 133 
Osmotic pressure, 5 
Oxalic acid, 102 
Oxidation, 30 

Perchloric acid, 97 
Permanganic acid, 126 
Phosphoric acid, 108 
Potassium, 96 
Precipitation, 20 
Preliminary tests, 127 

Reaction velocity, 13 
Reagents, 155 
Reduction, 30 
Reversible reaction, 4, 14 

Series, electrochemical, 59 
Silicic acid, 112 
Silver, 37 
Sodium, 96 

Solids, analysis of, 133 
Solubility-product, 19 
Solutions, analysis of, 131 
Spectrum analysis, 960 
Strontium, 90 
Sulphocyanic acid, 121 
Sulphuric acid, 100 
Sulphurous acid, 104 
Systematic analysis, 127 



INDEX 


Tables, acid analysis, 153 
atomic weights, 167 
basic analysis, 147 
percentage ionization, 165 
periodic system, 166 
reagents, 155 
solubilities, 160, 162 


Tartaric acid, 1x4 
Thiosulphuric acid, 106 
Tin, 60 

Valence, 10 

Zinc, 77 


171 






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